Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years ago

# 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
• 2 years ago

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.

• 2 years ago

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

• nbsale
Lv 6
2 years ago

69999 has sum 42=6x7

70000 has sum 7

That's the smallest number.

So you need 5 digits.

Others have the proof.