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# Given two consecutive numbers such that the sum of the digits of each number is a multiple of 7, ...?

Given two consecutive numbers such that the sum of the digits of each number is a multiple of 7. At least how many digits does the smaller number have?

A. 3

B. 4

C. 5

D. 6

E. 7 Relevance

The smaller number must end with some number of 9s. Otherwise, the digit sums will be 1 apart and both sums can't be multples of 7.

Take a look at what happens to the digit sum as you add 1 to some number of 9s:

1 + 9 = 10 : Digit sum goes from 9 to 1; a difference of 8 = 7 + 1

1 + 99 = 100 : Sum goes from 18 to 1; difference is 17 = 14 + 3

1 + 999 = 1,000 : Sum goes from 27 to 1: difference 26 = 21 + 5

See the pattern? With k low order 9s you get a difference of 7k plus the kth odd number. The next choice then has four 9s with a difference of 28 + 7 = 35, and that's the smallest number of trailing 9s that will produce a difference of that's a multiple of 7.

The larger number must end with four zeroes. The smallest number fitting that pattern that's a multiple of 7 is 70,000.

So, the smallest two numbers that work are 69,999 with a digit sum of 42, and 70,000 with a digit sum of 7.

The smaller number must have at least 5 digits.

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• Focus on the ones digit first.

Obviously if we just go from abc..4 to abc..5, the digit sum will differ by 1 we need them to differ by a multiple of 7.

So we must be going from abc..9 to abd..0. Also notice that we have a carry.

Next focus on the larger number. Since the 0 adds nothing to the digit sum the digits before the 0 must have a digit sum of 7. The smallest possible two-digit candidate would be 70, but 69 has a digit sum of 15 (1 more than a multiple of 7)

Switching to a three-digit sum, with the same pattern, it has to be 700 and 699. This time 699 has a digit sum of 24 (3 more than a multiple of 7).

If you see the pattern, each time we add one more digit, the remainder goes up by 2 (from 1 more, 3 more, 5 more, ...). At this point you should see that we need to go up 2 more digits so that we get it to 7 making the number a multiple of 7. To confirm this, let's just try the next candidates with 4-digits:

6999 and 7000 --> 6999 (digit sum = 33, 5 more than a multiple of 7).

Finally, 69999 and 70000 --> 69999 (digit sum = 42, 0 more than a multiple of 7).

That's the smallest possible answer with 5 digits.

C. 5

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• 69999 has sum 42=6x7

70000 has sum 7

That's the smallest number.

So you need 5 digits.

Others have the proof.

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• At first that seemed impossible. If you take a number and add 1, you just change the last digit, no matter how many digits it has, right? For instance, what's the next number after 42615? It's 42616. If the only digit that changes is the last one, and that digit changes by only 1, then the sum of the digits only changes by 1. Both sums can't be multiples of 7.

Aha, but what if the last digit is a 9? Look at 49 and 50. The last digit was reduced by 9 and the first digit increased by 1. The sum of digits changed by -9 + 1 = -8.

And if you have 3 digits, such as 199 and 200, then the sum of digits changes by -9 - 9 + 1 = -17.

Keep experimenting with that, till the change is a multiple of 7.

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