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 |
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together | 30 | + 1 = 16 times. |
2 |
Then, 13a x 13b = 2028
⟹ ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
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 |
= 540 + 8
= 548.
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.
∴ Larger number = (23 x 14) = 322.
= H.C.F. of 1651 and 2032 = 127.
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