Divisibility by 7
Number Theory
Level
3
Is the large number below divisible by 7?
This can only be determined by dividing by 7
No
Yes
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1 solution
Let’s tackle first the last six digits: 382,382
1 mod 7 =1; 10 mod 7 = 3; 100 mod 7 = 2; 1000 mod 7 = 6; 10,000 mod 7 = 4; 100,000 mod 7 = 5
If 10 is 7+3 then 80 is 8 ( 7 + 3 ) = 8 × 7 + 8 × 3 the first part is divisible by 7, so only the second part is of interest. Now apply this to all of them
3......8......2......3......8......2
5......4......6......2......3......1
Doable but complicated. We'd have to multiply the columns, then add the results to see if the sum is divisible by 7. But let's see what happens if we allow negative "remainders."
3......8......2......3......8......2
-2....-3....-1......2......3......1
-6...-24...-2....+6....+24..+2..=..0
Okay, so this one is divisible by 7. The same applies to the first six digits. 309,309. The remainders above repeat for higher powers of 10, endlessly. And being multiplied by a power of 10 doesn't change divisibility. The first 281 also cancels the second 281. They are far apart, but distance is such that the first is multiplied by the negatives -2, -3, -1, while the other is multiplied by the positives, 2, 3, 1. The first 672 likewise cancels the second 672, this time the first is multiplied by positives, the second by negatives.
So we are left with just 495,817. This one we'll do step by step.
4......9......5......8......1......7
-2....-3....-1......2......3......1
-8...-27...-5...+16...+3...+7..=.. -14
-14 is divisible by 7, therefore the entire 30-digit number is divisible by 7.