Sum Of Digits Of A Big Number

Number Theory Level 1

Find the sum of the digits of $\large 2^{2999} × 5^{3002}$ when multiplied out.

2^5

8 1 9

2 solutions

Dev Sharma
Oct 7, 2015

$2^{2999} × 5^{3002}$

= $2^{2999} × 5^{2999} × 5^3$

= $10^{2999} × 5^3$

= $125 × 10^{2999}$

Sum of digits : 1 + 2 + 5 = 8

Nice solution

Anuj Yadav - 5 years, 7 months ago

Awesome bro

Santosh Narva - 5 years, 7 months ago

Same.

Hey!! You should edit the second line.

Akshat Sharda - 5 years, 8 months ago

Realy!! RMO level...i doubt:((

Mohit Gupta - 5 years, 8 months ago

ADIPOLI(FANTASTIC)

Jerin Joy - 5 years, 8 months ago

awesome Soln

Madhulika Malhotra - 5 years, 7 months ago

This is a very strange problem from the point-of-view of an engineer and a mathematician, which I am both. It is neither a practical problem nor a thought-provoking one. Even in some field of applied math like cryptography, this problem is useless because it is much too easy.

Dale Wood - 4 years, 8 months ago

what solution is this

Jophin Joseph - 5 years, 8 months ago

What happen to the 1 with 2999 zeros? It should be 1+1+2+5 =9

rhey nevado - 5 years, 8 months ago

What happens when you multiply 125 by 1?

James Moors - 5 years, 8 months ago

125*10^2999 is 125 followed by 2999 0's. there is no other "1". so 1+2+5+0+0+0+0+0+0...=8

Lilu Barsanjaya - 5 years, 8 months ago

Now I get it

Maarten van Helden - 3 years ago

whats happened with 10 ^2999 Also statement is not clear

Zunair Nawaz - 5 years, 8 months ago

Sadasiva Panicker
Oct 12, 2015

2^2999 x 5^3002 = 2^2999 x 5^2999 x 5^3 =(2 X 5)^2999 x 125 = 10^2999 x 125 Therefore sum of digits= 1+2+5 = 8

Sum Of Digits Of A Big Number

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