∴ 13p + 11 = 17q + 9
⟹ 17q - 13p = 2
⟹ q = | 2 + 13p |
17 |
The least value of p for which q = | 2 + 13p | is a whole number is p = 26 |
17 |
∴ x = (13 x 26 + 11)
= (338 + 11)
= 349
(2n + 2)2 = (2n + 2 + 2n)(2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4.
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
264 → 11,3,4 (/)
396 → 11,3,4 (/)
462 → 11,3 (X)
792 → 11,3,4 (/)
968 → 11,4 (X)
2178 → 11,3 (X)
5184 → 3,4 (X)
6336 → 11,3,4 (/)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
= | 100 | x (1 + 100) - | 50 | x (1 + 50) |
2 | 2 |
= (50 x 101) - (25 x 51)
= (5050 - 1275)
= 3775.
Their number is 15
This is an A.P. in which a = 2, d = (4 - 2) = 2 and l = 30.
Let the number of terms be n. Then,
tn = 30 ⟹ a + (n - 1)d = 30
⟹ 2 + (n - 1) x 2 = 30
⟹ n - 1 = 14
⟹ n = 15
∴Sn = | n | (a + l) | = | 15 | x (2 + 30) = 240. |
2 | 2 |
Largest 5-digit number = 99999 91) 99999 (1098 91 --- 899 819 ---- 809 728 --- 81 --- Required number = (99999 - 81) = 99918.
Let the number of terms be n. Then,
a + (n - 1)d = 30
⟹ 2 + (n - 1) x 2 = 30
⟹ n = 15.
∴Sn = | n | (a + l) | = | 15 | x (2 + 30) = (15 x 16) = 240. |
2 | 2 |
47619 x 7 = 333333.
= (37 x 8q + 37 x 2) + 1
= 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.
4 | x y = (5 x 1 + 4) = 9 -------- 5 | y -1 x = (4 x y + 1) = (4 x 9 + 1) = 37 -------- | 1 -4 Now, 37 when divided successively by 5 and 4, we get 5 | 37 --------- 4 | 7 - 2 --------- | 1 - 3 Respective remainders are 2 and 3.
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