Consecutive Numbers Yielding Consecutive Digits

Algebra Level 3

Calvin chooses ten consecutive two-digit numbers and calculates their sum, which has consecutively increasing digits (for example, 567).

Find the greatest of those ten consecutive two-digit numbers.

Inspired by one of my problems.


The answer is 39.

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Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Yatin Khanna
Nov 15, 2016

Let the first number be x x ; then the sum of the 10 numbers = x + ( x + 1 ) + ( x + 2 ) . . . . . . + ( x + 9 ) = 10 x + 45 = x + (x+1) + (x+2) ...... + (x+9) = 10x + 45
Hence we know that the last digit of the sum is 5 5 .

Since the sum is consecutive increasing digits, thus it must be 345 345 .
Solving 10 x + 45 = 345 ⟹ x = 30 10x + 45 = 345 \implies x = 30 .
Hence, the greatest number is x + 9 = 30 + 9 = 39 x+9 = 30+9 =\boxed {39}

7 Helpful 1 Interesting 1 Brilliant 1 Confused

Great observation about the last digit! That immediately tells us that the sum must be 345.

I've tidied up the solution slightly to make it easier to read through.

Calvin Lin Staff - 4 years, 7 months ago

I missed the basic logic that finding one of the digits would completely define the sum. Thanks for the solution.

Nihar Mahajan - 4 years, 6 months ago
Nihar Mahajan
Nov 14, 2016

We know the formula for summing integers 1 to n as f ( n ) = n ( n + 1 ) / 2 f(n)=n(n+1)/2 .

Suppose that the greatest number is m. Then the smallest number among them would be (m-9). So their sum equals (sum from 1 to (m-10)) minus (sum from 1 to m). So we can write:

S = f ( m ) − f ( m − 10 ) S=f(m)-f(m-10) which on simplifying gives S = 10 m − 45 S=10m-45 .

Now, S = 100 a + 10 ( a + 1 ) + ( a + 2 ) = 10 m − 45 S=100a+10(a+1)+(a+2)=10m-45 which simplifies to 10 m = 111 a + 57 10m=111a+57 . Solving this Diophantine equation with a ≤ 9 a\leq 9 , we get a = 3 a=3 and plugging in its value yields the answer as 39 \boxed{39}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely done.

For the first half, it is easier to explain that m + ( m − 1 ) + ( m − 2 ) + … + ( m − 9 ) = 10 m − 45 m + (m-1) + (m-2) + \ldots + (m-9) = 10m - 45 .

If you look at solving the latter half, you actually do not need your second condition. IE we want to solve 10 m − 45 = 111 a + 12 10 m - 45 = 111 a + 12 , or that 10 m = 111 a + 57 10m = 111a + 57 . Hence we must have a = 3 a = 3 . I recommend making this edit to the problem.

Calvin Lin Staff - 4 years, 7 months ago

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