Five numbers are listed, one of them eleven digits long. From the right, add the digits in odd places and those in even places. If the two sums differ by a multiple of 11, so is the number. Pick out every multiple of 11.
Your answer
Watch outCounting from the left is natural, but places must be counted from the right.
How to solve
StrategyMake a Systematic List — There are exactly five candidates and one fixed procedure, so the safe way through is a list: for each number write the digits out from the right, mark which position each one sits in, total the odd-position digits, total the even-position digits, subtract. Almost every mistake in this kind of question is a mis-labelled position — reading from the left instead of the right, or losing track in the eleven-digit number — so writing the positions down beats doing it in my head. Before running the list I check the rule on a two-digit and a three-digit multiple of 11, which both shows me why the rule works and gives me the confidence to trust it on a number I could never divide by hand.
1STEP 1
Set up the positions, counting from the right
Number the places from the right-hand end.
5283: 5₄ 2₃ 8₂ 3₁
Reading a multi-digit number place by place is a Grade 4 skill; the only new instruction is to start the count at the right-hand end rather than the left.
4.NBT.A.2Make A Systematic List
2STEP 2
See why the alternating sum works, on small numbers first
Each place being ten times the one to its right is Grade 5 place value; noticing that every power of 10 is one away from a multiple of 11 is what makes the alternating +,-,+,- pattern appear.
5.NBT.A.1Solve An Easier Related Problem
The alternating sum works for 11 because the place values swing between one more and one less than a multiple of 11.
Why?
Ten is one less than 11 and a hundred is one more than a multiple of 11, and that pattern keeps repeating up the places.
🧱Divisibility tests come from place valueBecause 10, 100 and 1000 sit in a fixed relation to a divisor, only certain digits matter — the last two decide 4, the digit sum decides 3 and 9.
Why?
Because the sign flips with every step to the left, whether a digit is added or subtracted depends only on whether its place is odd or even.
🧱Even and odd behave predictablyTwice any whole number is even and adding an odd amount always swaps even with odd, so a total's parity is settled by its parts before anything is computed.
3STEP 3
(A) 5283
5283 differs by 8, so no.
(3+2)=5, (8+5)=13, 13-5=8 → not a multiple of 11
Two single-digit additions and one subtraction settle a four-digit question — far less work than dividing 5283 by 11.
4.NBT.B.4Make A Systematic List
4STEP 4
(B) 40964 — passes
40964 differs by 11, so yes.
(4+9+4)=17, (6+0)=6, 17-6=11 → multiple of 11
Recognising 11 itself as a multiple of 11 is the point of step 4 of the test — the difference does not have to be 0, it just has to be a multiple.
4.OA.B.4Make A Systematic List
5STEP 5
(C) 176234
176234 differs by 3, so no.
(4+2+7)=13, (3+6+1)=10, 13-10=3 → not a multiple of 11
With six digits the two groups have three digits each; keeping them in two separate columns as you read leftwards stops the digits getting mixed up.
4.NBT.B.4Make A Systematic List
6STEP 6
(D) 649325
649325 differs by 5, so no.
(5+3+4)=12, (2+9+6)=17, 17-12=5 → not a multiple of 11
Here the even-position sum is the bigger one; subtracting the smaller from the larger is fine, because only the size of the gap decides the answer.
4.NBT.B.4Make A Systematic List
7STEP 7
(E) 86629574592 — passes
86629574592 differs by 11, so yes.
(2+5+7+9+6+8)=37, (9+4+5+2+6)=26, 37-26=11 → multiple of 11
Eleven digits, but only eleven one-digit additions and one subtraction — the amount of work grows with the number of digits, not with the size of the number.
4.OA.B.4Make A Systematic List
8STEP 8
Confirm the two winners by actually dividing
Dividing confirms only those two come out exactly.
11 × 3724=40964, 11 × 7875415872=86629574592
The rejected numbers show what the test is really measuring: take the odd-position sum minus the even-position sum keeping the sign, and that number is off from the true remainder by a whole number of 11s. For 176234 it is 13-10=3 and the remainder is 3; for 649325 it is 12-17=-5 and the remainder is -5+11=6; for 5283 it is 5-13=-8 and the remainder is -8+11=3.
6.NS.B.2Look For A Pattern
Answer
40964, 86629574592
17 − 6 = 11, 37 − 26 = 11
Two out of five candidates passing is a believable hit rate for a hand-picked list — a random collection would be expected to yield about one in eleven, so the problem clearly seeded a couple of multiples on purpose. Both passing numbers gave a difference of exactly 11, not 0 and not 22, which is what you would expect from digits of this size: with five or six digits per group the two sums cannot differ by very much. Every alternating-sum verdict was confirmed by direct division: 40964 and 86629574592 divide by 11 exactly, while 5283, 176234 and 649325 leave remainders 3, 3 and 6 — and each of those remainders is exactly the signed alternating sum (-8, 3, -5) adjusted by a multiple of 11, so the three rejections are not just 'not multiples' but correctly rejected for the right amount.
Takeaway
Every place value is only one step away from a multiple of 11, so the digits alternately add and subtract — and an eleven-digit number becomes eleven easy additions.
Set up the positions, counting from the right
See why the alternating sum works, on small numbers first