The hcf and product of two numbers are 15 and 6300 respectively. the number of possible pairs of the numbers is
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Answered by
8
15a and 15b are our no.
where a and b are co prime no
15×15×ab= 6300
ab = 28
a. b.
4. 7. ( only co prime factor )
no are
15×4=60
15×7= 105
where a and b are co prime no
15×15×ab= 6300
ab = 28
a. b.
4. 7. ( only co prime factor )
no are
15×4=60
15×7= 105
Answered by
9
Answer:
2
Step-by-step explanation:
Let the two nos. be 15x & 15y.
A/Q 15x X 15y = 3600
xy = 3600/15x15= 28
Co-primes with product 28 are (28, 1) & (7, 4).
So, the possible nos. are (28x15, 1x15) & (7x15, 4x15).
i.e, (320, 15) & (105, 60).
Hence, no. of such pairs is 2.
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