Ooooo! That number's HUGE!!

All the 2 2 -digit numbers from 19 19 to 93 93 are written consecutively to from the number N = 19202122...93 N= 19202122...93 . Find the largest power of 3 3 that divides N N .

27 81 3 9

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3 solutions

Finding the sum of the numbers 19 to 93 using the formula for the sum of an arithmetic series, we get 1425. The prime factorization of it is 3 × 5 2 × 19 3 \times 5^{2} \times 19 . Since we are looking for the largest power of 3 that divides N , the answer is: 3 \boxed{3} .

Sum of arithmetic series won't be same as the sum of individual digits.

Kumar Shashwat - 6 years, 6 months ago
U Z
Oct 20, 2014

For 3 the divisibility test is that the sum of the digits should be divisible by 3

therefore

let s = 19 + 20 + 21 + 22 + . . . . . . . . . . . . . . . . . . . . . . . . . . . . + 93 s = 19 + 20 + 21 +22 + ............................ + 93

= 1 + 2 + . . . . . . . . . . . . + 18 + 19........... + 93 ( 1 + 2 + 3 + . . . . . . . . . + 18 ) = 1 + 2 +............+ 18 + 19 ........... + 93 - ( 1 +2 + 3 +......... + 18)

= 1581 171 = 1581 - 171

= 1410 = 1410

once again

1 + 4 + 1 + 0 1 + 4 +1 +0 is divisible by 3 and not by 9 , 27 or 81

therefore answer = 3

Here we don't have to add numbers. we need to add the individual digits. Like 19 should be added as 1+9=10 instead of 19

Kumar Shashwat - 6 years, 6 months ago
Aaaaa Bbbbb
Oct 18, 2014

192021...93 = 19 1 0 2 74 + 20 1 0 2 72 + . . . + 92 1 0 2 + 93 192021...93 = 19*10^{2*74}+20*10^{2*72}+...+92*10^2+93 = 9 A + 3 1400 = 3 ( 3 A + 1400 ) =9A+3*1400=3*(3A+1400) R e s u l t = 3 Result=\boxed{3}

PLEASE WORK OUT IN DETAIL.

Prabir Chaudhuri - 6 years, 7 months ago

Can you explain what the " 9 A + 3 × 1400 9A + 3 \times 1400 " is?

Calvin Lin Staff - 6 years, 7 months ago

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