Find HCF of x3 +c x2 -x + 2c and x2 + cx - 2 by long division method and get remainder
? Remainder = 2c - 2/c
Since,remainder should be zero
? 2c2 - 2 = 0
? 2(c2 -1) = 0 ? c= ± 1
Suppose two numbers are 3N and 5N.
Then, 3N x 5N = HCF x LCM
? 15N2 = 16 x 240
? N2 = 256
? N = 16
So, the number are 3N = 3 x 16 = 48 and 5N = 5 x 16 = 80.
The HCF of m and n is 1, so m and n are prime number.
Let m = 7 and n = 5 ? m + n = 12
HCF of(m + n)and m = HCF of 12 and 7 = 1
Similarly, HCF of (m - n) and n = 1
LCM of the number is always divisible by the HCF of the same numbers.
So, 78, 104 and 234 are all divisible by 26. Whereas, 144 is not divisible by 26.
Thus, 144 cannot be the LCM of the number whose HCF is 26.
LCM of 5, 10, 15, 20, 25 and 30 is 300. So, the bell will toll together after every 300s (5min).
So, the number of times they toll together = 60/5 + 1=13
Required number = H.C.F of (1657 - 6) and (2037 - 5)
= H.C.F of 1651 and 2032
= 127
The required greatest number is the HCF of (263 - 7, 935 - 7, 1383 - 7)
i.e., HCF of 256, 928 and 1376 = 32
The HCF of [(480 - 390), (620 - 480), (620 - 390)]
= HCF of 90, 140, 230 = 10
LCM of (12,18, 36) = 36
? A number greater than 300 on dividing by 36 leaves a remainder 4 = (36 x 9) + 4 = 328
Next numbe r = 328 + 36 = 364
? Sum of number = 328 + 364 = 692
132 = 2 x 2 x 3 x 11
240 = 2 x 2 x 2 x 2 x 3 x 5
228 = 2 x 2 x 3 x 19
Hence, HCF of 132, 204 and 228 = 2 x 2 x 3 = 12
We know that,
HCF of fractions = (HCF of numerators) / (LCM of denominators)
? Required HCF = (HCF of 4 and 7) / (LCM of 5 and 15)
=1/15
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