This year again

Level pending

$A$ is a number with $2015$ digits such that :

$A=\color{#D61F06}{2}\color{#69047E}{0}\color{#20A900}{1}\color{#3D99F6}{5}\color{#D61F06}{2}\color{#69047E}{0}\color{#20A900}{1}\color{#3D99F6}{5}\color{#D61F06}{2}\color{#69047E}{0}\color{#20A900}{1}\color{#3D99F6}{5}\color{#D61F06}{2}\color{#69047E}{0}\color{#20A900}{1}\color{#3D99F6}{5}...$ .

Find the maximum number of digits to be removed so that the sum is $2015$

The answer is 1612.

1 solution

Paola Ramírez
Mar 28, 2015

$\frac{2015}{5}=403$ so as minimum you can use $403$ digits five for complete the sum.

Now, in $A$ are $\frac{2015}{5}=403$ digits five $\therefore$ as maximum we can remove $2015-403=\boxed{1612}$ digits.

Here, realizing that using the 5's for the sum will fetch us maximum removal of digits was important!

Nihar Mahajan - 6 years, 2 months ago

good question as well as the solution.

abhigyan adarsh - 6 years, 1 month ago

This year again

The answer is 1612.

1 solution

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