Taking the sum of the digits, we have :
S1 = 9, S2 = 12, S3 = 18, S4 = 9, S5 = 21, S6 = 12, S7 = 18, S8 = 21, S9 = 15, S10 = 24.
Clearly, S2, S5, S6, S8, S9, S10 are all divisible by 3 but not by 9.
So, the number of required numbers = 6.
Let the two consecutive odd integers be (2x + 1) and (2x + 3)
Then, (2x + 3)2 - (2x + 1)2 = (2x + 3 + 2x + 1) (2x + 3 - 2x - 1) = (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8
Since 653xy is divisible by 5 as well as 2, so y = 0.
Now, 653x0 must be divisible by 8.
So, 3x0 must be divisible by 8. This happens when x = 2
x + y = (2 + 0) = 2
256256 = 256 x 1001; 678678 = 678 x 1001, etc.
So, any number of this form is divisible by 1001
-1 ≤ a; so a = -1
b ≤ 3; so b = 3
Now, 2a – 3b = -2 - 9 = -11
Therefore, the least value is -11
987 = 3 x 7 x 47.
So, required number must be divisible by each one of 3, 7, 47.
None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681
One-quarter of one-seventh = 1/4 x 1/7
= 1/28
Now, 1/28 of a land costs = Rs. 30,000
Thus, 8/35 of the land will cost = 30,000 x 28 x 8/35
= Rs. 1,92,2000.
90 = 10 x 9
Clearly, 653ab is divisible by 10, so b = 0
Now, 653a0 is divisible by 9.
So, (6 + 5 + 3 + a + 0) = (14 + a) is divisible by 9. So, a = 4.
Hence, (a + b) = (4 + 0) = 4.
20 x a = (64 + 36)(64 - 36) = 100 x 28
a = $\dfrac{100 x 28}{20}$ = 140
For every natural number n,$ (y^{n} - b^{n})$
is completely divisible by (y - b).