101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
= 540 + 8
= 548.
7 | = | 70 | ; | 13 | = | 65 | ; | 31 | = | 62 |
8 | 80 | 16 | 80 | 40 | 80 |
Since, | 70 | > | 65 | > | 63 | > | 62 | , so | 7 | > | 13 | > | 63 | > | 31 |
80 | 80 | 80 | 80 | 8 | 16 | 80 | 40 |
So, | 7 | is the largest. |
8 |
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = | ❨ | 551 | ❩ | = 19; Third number = | ❨ | 1073 | ❩ | = 37. |
29 | 29 |
∴ Required sum = (19 + 29 + 37) = 85.
Then, 37a x 37b = 4107
⟹ ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
∴ Greater number = 111.
Then, a + b = 55 and ab = 5 x 120 = 600.
∴ The required sum = | 1 | + | 1 | = | a + b | = | 55 | = | 11 |
a | b | ab | 600 | 120 |
∴ Larger number = (23 x 14) = 322.
= H.C.F. of 1651 and 2032 = 127.
Other number = | ❨ | 11 x 7700 | ❩ | = 308. |
275 |
(469 + 174)2 - (469 - 174)2 | =? |
(469 x 174) |
Given exp. = | (a + b)2 - (a - b)2 |
ab |
= | 4ab |
ab |
= 4 (where a = 469, b = 174.)
Then x + 5x + 10x = 480
⟹ 16x = 480
∴ x = 30.
Hence, total number of notes = 3x = 90.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.