Problem
Append digits to hit several multiples
Three more digits are written after 527, making a six-digit number. It must be a multiple of 3 and a multiple of 4. It must be a multiple of 5 as well. Find the smallest number that works.
Reasoning · Grade 5-1 Divisibility Tests
Your answer
Watch out Multiple of 5 allows 0 or 5, but with 4 in play only 0 survives.
How to solve
Strategy Eliminate Possibilities — There are only 1000 candidates (527000 to 527999), and each divisibility test wipes out most of them, so I apply the tests in the order that cuts the hardest. The test for 5 pins the last digit down to one of two values. The test for 4 then only involves the last two digits. The test for 3 only involves the digit sum. Applying them from the most restrictive to the least leaves a handful of numbers, and I take the smallest. Because 'smallest' means the hundreds digit matters most, I try A = 0 first and only move up if nothing works.
1STEP 1
Use the multiple-of-5 test to fix the ones digit
Multiple of 5 makes the ones digit 0 or 5.
527ABC is a multiple of 5 → C=0 or C=5
Counting by fives always lands on a number ending in 0 or 5, so one glance at the last digit settles it - no dividing needed.
4.OA.B.4Eliminate Possibilities2STEP 2
Use the multiple-of-4 test to fix the tens digit
Multiple of 4 needs the last two digits to be an even ten and 0.
B0 ∈ {00, 20, 40, 60, 80} → B ∈ {0,2,4,6,8}
Because 100 is itself a multiple of 4, everything above the tens place is already divisible by 4 and can be ignored - only the last two digits decide, which is why this test is so quick.
4.OA.B.4Eliminate Possibilities3STEP 3
Use the multiple-of-3 test on the digit sum
Multiple of 3 needs the digit sum to be a multiple of 3.
5+2+7+A+B+0 = 14+A+B ≡ 0 (mod 3) → A+B ∈ {1,4,7,10,13,16}
The digit-sum rule works because 10, 100, 1000 and so on are each one more than a multiple of 3, so every digit contributes just itself to the remainder - which is why adding the digits is enough.
4.OA.B.4Look For A PatternA number is a multiple of 3 exactly when its digits add to a multiple of 3, so the test needs no dividing.
Why?
Ten, a hundred and a thousand are each one more than a multiple of 3, so every digit contributes only itself to the remainder.
🧱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?
Every candidate whose digit sum fails is struck out at once, so the search shrinks before any long division starts.
🧱Rule out the rest and the last one is forcedIf every possibility but one has been ruled out, the survivor must be the answer — no further checking is needed.
4STEP 4
Search for the smallest by trying the hundreds digit from 0 upward
Raising the hundreds digit from 0 gives 527040.
A=0 → B=4 → 527040
Comparing six-digit numbers is place-value work from Grade 4: the leftmost digit that can still change is the hundreds digit, so making it as small as possible beats every other choice no matter what follows.
4.NBT.A.2Make A Systematic List5STEP 5
Check 527040 against all three conditions
It passes all three tests, so the answer is 527040.
527040 ÷ 3 = 175680, 527040 ÷ 4 = 131760, 527040 ÷ 5 = 105408
Because 3, 4 and 5 share no common factor, being a multiple of all three is the same as being a multiple of 3 x 4 x 5 = 60 - and 527040 = 60 x 8784, which is a single check that covers all three at once.
6.NS.B.4Guess And CheckAnswer
527040
527040 ÷ 3 = 175680
The answer must be a six-digit number starting with 527, and 527040 is. Being a multiple of 3, 4 and 5 at once is the same as being a multiple of 60, and the first multiple of 60 at or above 527000 is found by noting 527000 = 60 x 8783 + 20, so the next one is 527000 + 40 = 527040 - the same answer by a different route. Nothing between 527000 and 527040 can work, because those numbers would need a ones digit of 0 (multiple of 5) and only 527000 qualifies, whose digit sum 14 is not a multiple of 3.
Takeaway
Apply the strictest test first: the last digit had to be 0, then the tens digit had to be even, and only then did the digit-sum rule have almost nothing left to check.
- Use the multiple-of-5 test to fix the ones digit
- Use the multiple-of-4 test to fix the tens digit
- Use the multiple-of-3 test on the digit sum
- Search for the smallest by trying the hundreds digit from 0 upward
- Check 527040 against all three conditions