Ooooo! That number's HUGE!!
All the 2 -digit numbers from 1 9 to 9 3 are written consecutively to from the number N = 1 9 2 0 2 1 2 2 . . . 9 3 . Find the largest power of 3 that divides N .
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3 solutions
Sum of arithmetic series won't be same as the sum of individual digits.
For 3 the divisibility test is that the sum of the digits should be divisible by 3
therefore
let s = 1 9 + 2 0 + 2 1 + 2 2 + . . . . . . . . . . . . . . . . . . . . . . . . . . . . + 9 3
= 1 + 2 + . . . . . . . . . . . . + 1 8 + 1 9 . . . . . . . . . . . + 9 3 − ( 1 + 2 + 3 + . . . . . . . . . + 1 8 )
= 1 5 8 1 − 1 7 1
= 1 4 1 0
once again
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
1 9 2 0 2 1 . . . 9 3 = 1 9 ∗ 1 0 2 ∗ 7 4 + 2 0 ∗ 1 0 2 ∗ 7 2 + . . . + 9 2 ∗ 1 0 2 + 9 3 = 9 A + 3 ∗ 1 4 0 0 = 3 ∗ ( 3 A + 1 4 0 0 ) R e s u l t = 3
PLEASE WORK OUT IN DETAIL.
Can you explain what the " 9 A + 3 × 1 4 0 0 " is?
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 × 1 9 . Since we are looking for the largest power of 3 that divides N , the answer is: 3 .