Answer: Option D
Explanation:
Answer: Option B
Largest 5-digit number = 99999 91) 99999 (1098 91 --- 899 819 ---- 809 728 --- 81 --- Required number = (99999 - 81) = 99918.
This is an A.P. in which a = 51, l = 100 and n = 50.
= (25 x 151) = 3775.
Answer: Option C
The smallest 6-digit number 100000. 111) 100000 (900 999 ----- 100 --- Required number = 100000 + (111 - 100) = 100011.
35423 317 x 89 = 317 x (90 -1 ) + 7164 = (317 x 90 - 317) + 41720 = (28530 - 317) ----- = 28213 84307 - 28213 ----- 56094 -----
(12 + 22 + 32 + ... + 102) = ?
We know that (12 + 22 + 32 + ... + n2)
Putting n = 10, required sum
When n is even. (xn - an) is completely divisible by (x + a)
(17200 - 1200) is completely divisible by (17 + 1), i.e., 18.
(17200 - 1) is completely divisible by 18.
On dividing 17200 by 18, we get 1 as remainder.
Answer: Option A
For every natural number n, (xn - an) is completely divisible by (x - a).
Let x = 6q + 3.
Then, x2 = (6q + 3)2
= 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
Thus, when x2 is divided by 6, then remainder = 3.
(17200 - 1200) is completely divisible by (17 + 1), i.e., 18.
(17200 - 1) is completely divisible by 18. On dividing 17200 by 18, we get 1 as remainder.
