Question

Another ridiculously stupid question from the American Mathematics Competition.

Ximena lists the whole numbers 1 through 30 once. Emilio copies Ximena's numbers, replacing each occurrence of the digit 2 by the digit 1. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?

A. 13
B. 26
C. 102
D. 103
E. 110

Is there any easier way to do this instead of adding each number one by one?

Answer 1

Using the same formula as before,\[\large\rm \sum_{i=1}^{30}i=\frac{30(30+1)}{2}=465\]From there,
you can just pick out the numbers that have two's in them I guess.
2
12
20 - 29

We lose 1 from each 2 in the ONES PLACE.
We lose 10 from each 2 in the TENS PLACE.

So we're losing 1x2 in the ONES,
and 10x10 in the TENS, ya?

So 102 larger if I added those correctly.

Answer 2

Woops I forgot that 22 has two of those 2's in it! :)
So we lose another from the ONES PLACE.

103

Answer 3

Oh thanks for the explanation! So to find the sum of the integers to the integer n, the formula is n(n+1)/2

Answer 4

I didn't know a way to figure out a way to add whole numbers from 1 to 30, and I never knew there was a formula for that.