QuanticaPrealgebraOpen in the courseTake a number like 6,351. To find out whether 3 divides it you could do the long division, or you could add up the digits and know in a second. Every shortcut like that comes out of place value.
Problem
Five claim tags have numbers 2,470; 3,805; 6,118; 9,640; and 7,333. Without dividing, decide which are multiples of 5. Which single digit settles the question? How many of the five tags win a prize?
Show a hint
- You do not have to divide the whole tag by 5. Write one tag as a stack of tens plus its leftover ones digit, like 2,470 = 247 tens and 0 ones, and watch which part you can stop worrying about.
- A whole number of tens is already a multiple of 5, since every ten is 5 twice. By closure from the last lesson, throwing away a multiple of 5 leaves the question resting on the single ones digit. Which ones digits make the leftover a multiple of 5?
- Sort the tags by their last digit alone. A tag wins for 5 when it ends in 0 or 5, and it also wins for 10 only when it ends in 0.
Show the full solution
Read the last digit of each tag. \(2{,}470\) ends in 0, \(3{,}805\) ends in 5, and \(9{,}640\) ends in 0, so those three are multiples of 5, while \(6{,}118\) and \(7{,}333\) end in 8 and 3. That is \(\boxed{3}\) tags. The ones digit decides it alone, because everything above it is a stack of tens and every ten already holds a 5.
Problem
Band sections have 84,317; 2,000,006; 55,550; and 9,999 players. Which counts are even and can line up in perfect pairs? How many of the four can pair up perfectly?
Show a hint
- Pairing up two to a row works exactly when the count is even, so the real question is which of these four counts is even. Look hard at the end of each number before you reach for anything bigger.
- Try writing a number as its tens part plus its last digit, like \(55{,}550 = 55{,}55 \times 10 + 0\) or \(84{,}317 = 8{,}431 \times 10 + 7\). Any whole number of tens is already even, so the only thing still in doubt is that final leftover digit.
- For \(2{,}000{,}006\), all those zeros sit in the tens part and are already even on their own. Only the final \(6\) is left to decide the whole question, and \(6\) is even.
Show the full solution
Pairing up with nobody left over is just asking whether the count is even, so read the last digit. \(84{,}317\) ends in 7 and \(9{,}999\) ends in 9, both odd. \(2{,}000{,}006\) ends in 6 and \(55{,}550\) ends in 0, both even, so \(\boxed{2}\) sections pair up perfectly. Everything above the units digit is a whole number of tens, and tens are even, so only that last digit can tip a number to odd.
The last digit settled 2, 5, and 10 because every higher place is a multiple of ten. Now try 4. Ten is not a multiple of 4, so one digit is not enough. We need to cut earlier and keep a bigger tail. Where should the cut go?
Problem
A dealer splits stacks of 5,328; 7,914; 13,500; and 60,742 cards into complete hands of 4. Write each as hundreds plus its last two digits. How many stacks deal out evenly?
Show a hint
- Try splitting each deck at the hundreds, like \(7{,}914 = 79\times100 + 14\). One hundred is \(25\) full hands of \(4\), so every full hundred already deals out evenly. That means the hundreds part never causes trouble, no matter how big it is.
- If the hundreds part always deals out evenly, then whether the whole stack works comes down to that small leftover number in the last two digits. So only check whether \(28\), \(14\), \(00\), and \(42\) each split into hands of \(4\).
Show the full solution
One hundred is \(4\times 25\), so every full hundred deals out evenly and only the last two digits matter. $$5{,}328 = 53\times100 + 28,\qquad 7{,}914 = 79\times100 + 14$$ $$13{,}500 = 135\times100 + 00,\qquad 60{,}742 = 607\times100 + 42$$ The tail \(28 = 4\times 7\) works and the empty tail \(00\) works, while \(14\) and \(42\) are not multiples of 4. That leaves \(\boxed{2}\) stacks. Ten is not a multiple of 4, which is why one digit is not enough here and you have to keep two.
Problem
Arcade machines show totals 47,216; 9,360; and 128,500. Test only the last three digits of each for divisibility by 8. How many are divisible by 8?
Show a hint
- A whole number of strips means the total is a multiple of \(8\). Look at \(1000\). Can you write \(1000\) itself as a whole number times \(8\)? If every full thousand is already a stack of strips, then the thousands part can be thrown away and only what is left below a thousand can decide things.
- Split each total right after the last three digits, like \(47{,}216 = 47\times 1000 + 216\). Closure lets you drop the \(47\times 1000\) chunk because it is a pile of complete strips, so only the three-digit tail, \(216\), is left to test against \(8\).
- Now line the tests up. For \(2\) you read \(1\) digit, for \(4\) you read \(2\) digits, and for \(8\) you read \(3\). Ask which power of ten each of \(2, 4, 8\) first divides into cleanly, and the digit counts will line up by themselves.
Show the full solution
One thousand is \(8\times 125\), so every full thousand is a whole number of 8s and only the last three digits matter. For \(47{,}216\) the tail \(216 = 8\times 27\) fits, for \(9{,}360\) the tail \(360 = 8\times 45\) fits, and for \(128{,}500\) the tail \(500\) lands between \(8\times 62 = 496\) and \(8\times 64 = 512\), so it fails. That is \(\boxed{2}\) totals. You read as many digits as the first power of ten your divisor divides cleanly, so \(2 \mid 10\), \(4 \mid 100\), and \(8 \mid 1000\) give 1 digit, 2 digits, and 3 digits.
Try the same tail trick on 9. It would need a power of ten that 9 divides, and there is none, since \(10=9+1\), \(100=99+1\), and \(1000=999+1\). Every power of ten misses a multiple of 9 by exactly 1, and because the miss is the same size every time, we can still build a test out of it.
Problem
The stamp number is 5,742. Break it by place value, write each power of ten as a multiple of 9 plus 1, and let the multiples fall away. What leftover decides whether 9 divides 5,742?
Show a hint
- Do not divide the whole 5,742 at once. Split it by place value, \(5742 = 5\times 1000 + 7\times 100 + 4\times 10 + 2\), and look at each power of ten on its own.
- Write each power of ten as a multiple of 9 plus 1, so \(1000 = 999 + 1\), \(100 = 99 + 1\), \(10 = 9 + 1\). Every '\(+1\)' hands its digit straight back to you.
- After the \(999\), \(99\), and \(9\) chunks fall away as multiples of 9, the only piece left to check is \(5 + 7 + 4 + 2\).
Show the full solution
Spread \(5742\) out by place value and write each power of ten as a multiple of 9 plus 1. $$5742 = 5(999+1) + 7(99+1) + 4(9+1) + 2 = \big(5\times 999 + 7\times 99 + 4\times 9\big) + \big(5 + 7 + 4 + 2\big)$$ The first group is built from \(999\), \(99\), and \(9\), so closure drops it as a multiple of 9. What decides the question is the digit sum \(5 + 7 + 4 + 2 = \boxed{18}\). Since \(18 = 9\times 2\), the stamp number is divisible by 9, and \(5742 \div 9 = 638\) confirms it.
Problem
A view counter reads 5,839,206. Sum its digits, and if the result has more than one digit, sum again. What single digit do you land on, and what does it say about divisibility by 3 and by 9?
Show a hint
- Add the seven digits of 5,839,206 together. You should get a two-digit number, so do not stop there.
- Your first total is 33. Now add ITS digits, 3 + 3, to crush it down to one digit.
- The single digit you reach is the remainder when 5,839,206 is divided by 9. Compare that digit to 3 and to 9 to read off which one divides the count.
Show the full solution
Add the digits, \(5+8+3+9+2+0+6 = 33\). That is still two digits, so add again, \(3+3=\boxed{6}\). A number and its digit sum leave the same remainder under 9, so this single digit is that remainder. It is a multiple of 3 but not of 9, so \(5{,}839{,}206\) is divisible by 3 and not by 9.
Try 11 next. We have \(10=11-1\), so ten leaves remainder \(-1\). Then \(100=99+1\) leaves \(+1\), and \(1000=1001-1\) leaves \(-1\). The remainders flip between \(-1\) and \(+1\) as you climb the places, so the digits will need alternating signs.
Problem
Cards read 90,728 and 53,481. Exactly one is a multiple of 11. Since \(10\equiv-1\pmod{11}\), combine digits with alternating signs. Which card is the multiple?
Show a hint
- Start at \(10 = 11 - 1\). What does the next place value, \(100\), sit near? Try writing \(100\) as a multiple of \(11\) plus a little leftover, and then do the same for \(1000\), watching the leftover flip sign each time.
- Since the leftovers go \(+1, -1, +1, -1, \ldots\) from the units place upward, a number leaves the same remainder against \(11\) as its digits combined with alternating signs, starting \(+\) at the units. Build that alternating sum for each card.
- For \(90{,}728\) the alternating sum from the units is \(8 - 2 + 7 - 0 + 9\). Work that out, do the same for \(53{,}481\), and check which result is itself a multiple of \(11\).
Show the full solution
Since \(10 = 11 - 1\), \(100 = 9\times 11 + 1\), and \(1000 = 91\times 11 - 1\), the place values carry signs \(+, -, +, -\) from the units up once closure throws the multiples of 11 away. Combine each card's digits with those signs. $$8 - 2 + 7 - 0 + 9 = 22, \qquad 1 - 8 + 4 - 3 + 5 = -1$$ Only \(22 = 2\times 11\) is a multiple of 11, so the card is \(\boxed{90{,}728}\). The alternating sum gives the remainder against 11, and \(53{,}481\) misses by exactly that \(-1\).
So far we can test 2, 3, 4, 5, 8, 9, 10, and 11, but not 6, 12, or 15. Each of those factors into divisors we already know how to test, so it is tempting to run both piece tests and combine the results. That does work, but only for certain choices of the two pieces.
Problem
A clerk thinks: if a pile is even and divisible by 6, it is divisible by 12. Find one pile divisible by 2 and by 6 that is NOT divisible by 12.
Show a hint
- Test the clerk's two claims on a small count first. Pick a number you already know fills whole cartons of 6, then check both whether it is even and whether it splits into crates of 12.
- The trap is that 6 already carries a factor of 2 inside it, so being even tells you nothing new beyond what divisibility by 6 already promised. You want a multiple of 6 that is missing a second factor of 2.
- Multiples of 6 go 6, 12, 18, 24, 30, and so on. Every other one is odd-times-6, which keeps just one factor of 2. Take the first such count after 6 itself.
Show the full solution
Take \(18\). It fills whole cartons since \(18 = 6\times 3\), and it is even since \(18 = 2\times 9\), so it passes both of the clerk's tests. But \(18 = 12 + 6\), and that leftover 6 is only half a crate, so 18 is not divisible by 12. The clerk is wrong, and \(\boxed{18}\) proves it. Splitting a divisor into two tests works only when the pieces are coprime. Here 6 already carries the 2, so \(12 = 2\times 6\) counts the same factor of 2 twice, while \(12 = 2^2\times 3\) needs two of them. The honest split is \(12 = 3\times 4\).
Problem
A print shop has 25,830 flyers. Using the matching shortcut for each of 2, 3, 4, 5, and 9, decide which stack sizes divide 25,830 evenly. How many of the five work?
Show a hint
- Each divisor has its own window into the number. For 2 and 5 you only need the last digit, for 4 the last two digits, and for 3 and 9 the digit sum. Look at the right part of 25,830 for each test instead of dividing the whole thing.
- The flyer count ends in 0, which settles 2, 5, and 4 in one glance, and the digits add to a tidy round total that settles 3 and 9 together. Add 2+5+8+3+0 and see what it is a multiple of.
- Four of the five fit. The only one to double check is 4, where ending in 0 is not enough on its own, so peek at the full last two digits, 30, and ask if that is a multiple of 4.
Show the full solution
The last digit is \(0\), so 2 and 5 both divide the run. For 4, read the last two digits, and \(30 = 4\times 7 + 2\) is not a multiple of 4, so 4 fails. For 3 and 9, read the digit sum. $$2+5+8+3+0 = 18 = 9\times 2$$ So 9 divides the run, and 3 does too since 3 divides every 9. That is \(\boxed{4}\) of the five stack sizes. Ending in 0 settles 2 and 5 but never 4, which needs the full last two digits.
Run the tests backward. Instead of judging a finished number, suppose one digit is hidden and the number must be divisible by something. The same place-value thinking applies, but now you solve for the missing digit rather than just checking.
Problem
A boarding pass prints the flight number as 46,N12, but the hundreds digit N is smudged. The airline only issues numbers divisible by 9. Use the digit-sum test backward to find N.
Show a hint
- The digit-sum test says a number is divisible by \(9\) exactly when the sum of its digits is a multiple of \(9\). Add up the digits you can see and leave \(N\) sitting in the sum.
- You should land on \(13+N\). Now ask which single digit you can drop in so that \(13+N\) becomes a multiple of \(9\). The next multiple of \(9\) past \(13\) is \(18\).
- Only one digit from \(0\) through \(9\) makes \(13+N\) reach a multiple of \(9\). Which one closes the gap up to \(18\)?
Show the full solution
Add the digits of \(46{,}N12\) and keep \(N\) in the sum. $$4+6+N+1+2=13+N$$ For divisibility by 9 this has to be a multiple of 9, and the only one a single digit can reach from 13 is 18. So \(13+N=18\) and \(N=\boxed{5}\). The flight number is \(46{,}512\), whose digits sum to \(18 = 2\times 9\). The digit-sum test runs backward just as well as forward, which is often enough to recover a hidden digit.
Problem
A label reads 543N6 with the tens digit N smudged. It must pass both the sorter for multiples of 9 and the sorter for multiples of 4. What digit is N?
Show a hint
- Two machines means two tests, and each test watches a different part of the number. The multiple of 9 test cares about the whole digit sum, while the multiple of 4 test cares only about the last two digits, N6.
- Add the known digits, 5+4+3+6 = 18, so the digit sum is 18+N. For that to be a multiple of 9, N has just two choices. Now check which of those choices makes the last two digits a multiple of 4.
Show the full solution
For the 9 sorter, the digit sum is \(5+4+3+N+6 = 18+N\), which is a multiple of 9 only for \(N=0\) (sum 18) or \(N=9\) (sum 27). For the 4 sorter, only the last two digits matter, since 100 is a multiple of 4. With \(N=0\) the tail is \(06 = 6\), not a multiple of 4, but with \(N=9\) the tail is \(96 = 4\times 24\). Only one digit clears both machines, $$N = \boxed{9}.$$ Two tests read different pieces of the same number, so stacking them pins the hidden digit where the survivor lists overlap.
Every test worked the same way. Cut where one chunk becomes a multiple of the divisor, let closure drop that chunk, and read the small tail left behind. The cut lands in a different place for each divisor, and these tests will do real work in the prime factoring ahead.
Practice these ideas
Practice
Five raffle tickets have numbers 3,160; 8,002; 47,225; 90,118; and 6,500. Which are divisible by 5? How many win a prize?
Show the solution
Read the last digit of each ticket. \(3{,}160\) ends in 0, \(47{,}225\) ends in 5, and \(6{,}500\) ends in 0, so those three are divisible by 5, while \(8{,}002\) and \(90{,}118\) end in 2 and 8. That is \(\boxed{3}\) winners. Everything above the units digit is a multiple of ten, and ten is built from a 5, so only the last digit decides.
Practice
Five flight log codes are 1,316; 5,550; 23,128; 8,914; and 40,016. Which are divisible by 4? How many qualify?
Show the solution
One hundred is \(4 \times 25\), so every full hundred is a multiple of 4 and closure lets you keep only the last two digits. For \(1{,}316\) the tail is \(16 = 4\times 4\), for \(23{,}128\) it is \(28 = 4\times 7\), and for \(40{,}016\) it is \(16\) again, all passing. The tails \(50\) and \(14\) each sit two past a multiple of 4, so those fail. That leaves \(\boxed{3}\) codes.
Practice
A track code reads 49,d06. To keep it divisible by 3, what is the largest digit d can be?
Show the solution
The digit sum is \(4+9+d+0+6 = 19+d\), and divisibility by 3 needs that total to be a multiple of 3. Since \(19\) leaves a remainder of 1, \(d\) has to supply a remainder of 2, which happens for \(d = 2,\ 5,\ 8\). The largest single digit there is \(\boxed{8}\). With \(d=8\) the sum is \(27 = 3\times 9\), while \(d=9\) would give 28 and miss.
Practice
Batch codes are 5,431; 7,002; 12,345; 81,000; and 6,453. How many are divisible by 9?
Show the solution
Add the digits of each code, since a number and its digit sum leave the same remainder under 9. The five sums are \(5+4+3+1=13\), \(7+0+0+2=9\), \(1+2+3+4+5=15\), \(8+1+0+0+0=9\), and \(6+4+5+3=18\). The multiples of 9 are \(9\), \(9\), and \(18\), belonging to \(7{,}002\), \(81{,}000\), and \(6{,}453\), so \(\boxed{3}\) codes qualify. The digit sum works because every power of ten is one more than a run of nines.
Practice
Every divisibility test reads only the piece the tossed chunk cannot hide. Write a number as digits times powers of ten, cut where one chunk becomes a multiple of the divisor, and closure throws that chunk away. The last digit settles 2, 5, 10, since the whole upper part is a multiple of all three. The last two digits settle 4 and the last three settle 8, since \(100 = 4 \times 25\) and \(1000 = 8 \times 125\). The digit sum settles 3 and 9, since each power of ten is one above a multiple of 9. The alternating digit sum settles 11, since the powers of ten land just below then above a multiple of 11, flipping sign. A composite settles by splitting it into coprime factors and testing each.
Practice
Five sticker orders are 4,128; 7,533; 90; 25,164; and 3,400. Which are divisible by 6? How many can the printer fill?
Show the solution
A number is divisible by 6 exactly when it is divisible by both 2 and 3. \(4{,}128\) is even with digit sum 15, \(90\) is even with digit sum 9, and \(25{,}164\) is even with digit sum 18, so those three pass both tests. \(7{,}533\) is odd, and \(3{,}400\) is even but its digit sum 7 is not a multiple of 3. The printer can fill \(\boxed{3}\) orders. The split \(6 = 2\times 3\) works because 2 and 3 share no common factor.
Practice
A bib number reads 34_7 with the tens digit hidden. Every bib is divisible by 9. Use the digit-sum test backward to find the hidden digit.
Show the solution
Call the hidden tens digit \(d\). The digit sum is \(3 + 4 + d + 7 = 14 + d\), and it must be a multiple of 9. Since \(d\) runs from 0 to 9, the sum runs from 14 to 23, and the only multiple of 9 in that window is 18, so \(14 + d = 18\) gives \(d = \boxed{4}\). That makes the bib \(3447\), and \(9 \times 383 = 3447\). The digit-sum test runs in reverse, turning a hidden digit into a small equation.
Practice
Daily krill counts are 3,128; 17,400; 9,990; 52,016; and 6,300. Which are divisible by 8? How many work?
Show the solution
A thousand is \(8\times 125\), so the thousands part is always a multiple of 8 and only the last three digits matter. For \(3{,}128\) the tail is \(128 = 8\times 16\), for \(17{,}400\) it is \(400 = 8\times 50\), and for \(52{,}016\) it is \(016 = 8\times 2\), so those three work. For \(9{,}990\) the tail \(990\) sits 6 past \(8\times 123 = 984\), and for \(6{,}300\) the tail \(300\) sits 4 past \(8\times 37 = 296\), so both fail. That is \(\boxed{3}\) counts.
Practice
Among house numbers 1 through 500, how many end in 6 and are divisible by 4?
Show the solution
Divisibility by 4 reads only the last two digits, since \(100 = 4\times 25\) sweeps every full hundred away. Among the endings \(06, 16, 26, 36, 46, 56, 66, 76, 86, 96\), the multiples of 4 are \(16, 36, 56, 76, 96\), five per hundred. From 1 through 500 there are 5 full hundreds, so the total is \(5 \times 5 = \boxed{25}\). The winners are exactly the endings with an odd tens digit, which is why the same five repeat in every hundred.
Practice
A count reads 8_,386 with the thousands digit missing. It is divisible by 9. Find the hidden digit, then decide whether the full number is also divisible by 4.
Show the solution
The visible digits add to \(8+3+8+6 = 25\), so with the hidden digit \(d\) the digit sum is \(25 + d\). That runs from 25 to 34, and the only multiple of 9 in range is 27, so \(d = \boxed{2}\) and the count is \(82{,}386\). For 4, only the last two digits matter, and \(86 = 4 \times 21 + 2\), so \(82{,}386\) is not divisible by 4. Each test reads a different chunk, so passing one never drags the other along.
Practice
A stamp serial reads 7_,320 with the thousands digit smudged. It must be divisible by both 3 and 8. How many digits could hide under the smudge?
Show the solution
For 8, only the last three digits matter, and \(320 = 8 \times 40\), so the serial is divisible by 8 whatever hides in the thousands place. For 3, the digit sum is \(7 + \square + 3 + 2 + 0 = 12 + \square\), and since 12 is already a multiple of 3, the sum works exactly when \(\square\) is a multiple of 3, giving \(0, 3, 6, 9\). That is \(\boxed{4}\) digits. The smudge sits above the window the 8 test reads, so only the 3 test rules anything out.
Practice
Tickets read 2,051; 4,613; 5,000; 8,267; and 3,002. Unlike 2 or 5, the number 7 has no digit shortcut, so divide honestly. How many win a cookie?
Show the solution
Seven has no digit shortcut, so divide. $$2{,}051 = 7 \times 293, \qquad 4{,}613 = 7 \times 659, \qquad 8{,}267 = 7 \times 1{,}181$$ Those three land exactly, while \(5{,}000 = 7 \times 714 + 2\) and \(3{,}002 = 7 \times 428 + 6\) leave remainders. So \(\boxed{3}\) tickets win a cookie. Not every divisor comes with a digit test, and for one like 7 you divide and read the remainder.
Practice
A wristband serial reads 41_9 with the tens digit hidden. Every serial is divisible by 11. Use the alternating-sum test backward to find the hidden digit.
Show the solution
Call the hidden tens digit \(d\). Reading from the right with the units plus, the alternating sum is \(9 - d + 1 - 4 = 6 - d\), and it must be a multiple of 11. Since \(d\) runs from 0 to 9, \(6 - d\) runs from 6 down to \(-3\), and the only multiple of 11 there is 0, so \(d = \boxed{6}\). That makes the serial \(4169\), and \(11 \times 379 = 4169\). The alternating-sum test runs backward the same way the digit-sum test does.
Practice
Since \(1001=7\times11\times13\), chop the number into three-digit blocks from the right and alternately subtract and add them. The result shares the same divisibility by 7, 11, and 13.