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.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
∴ Required number = (90 x 4) + 4 = 364.
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30 ---------------------------- = 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15 ---------------------------- Required number = (1260 ? 2) | 2 - 3 - 7 - 5 = 630.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
∴ The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
∴ Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
∴ Required number = (840 x 2 + 3) = 1683.
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Then, a + b = 55 and ab = 5 x 120 = 600.
∴ The required sum = | 1 | + | 1 | = | a + b | = | 55 | = | 11 |
a | b | ab | 600 | 120 |
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.
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.
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 |
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