(52 - 33) = 19, (78 - 59) = 19 and (117 - 98) = 19
LCM of 52, 78 and 117 is 468.
Required least number = 468 + 19 = 487
Sum of digits of 487 = 4 + 8 + 7 = 19
(54 - 35) = 19, (80 - 61) = 19 and(119 - 100) = 19
LCM of 54, 80 and 119 is 257040.
Required least number = 257040 + 19 = 257059
Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28
Go through options.
Put m = 1 in two given expression,we have
x3 - 10x2 + 31x - 30 and x2 - 8x + 15
x2 - 8x + 15 = (x - 5) (x - 3)
x3 - 10x2 + 31x - 30 = (x - 2) (x - 5) (x - 3)
? (x - 5) (x - 3) = x2 - 8x + 15 is the HCF of both expressions.
? m = 1
Minimum number of rows = Maximum number of trees in each row
= HCF of 30,45 and 60 = 15
? Number of rows = (30 + 45 + 60)/15 = 9
HCF of 403,456 and 496 = 31
Now, the number of bottles required to put 403 L of petrol = 403/31 = 13
Similarly,the number of bottles required to put 456 L of diesel = 465/31 = 15
The number of bottles required to put 496 L of mobile oil = 496/31 = 16
? Least number of bottles required having 31 L as capacity = 13+15 + 16 = 44
The time which all the three persons meet will be the LCM of the time by each person individually to complete one round.
252 = 22 x 71 x 91
308 = 22 x 71 x 111
198 = 21 x 91 x 111
? LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772
So,A,B and C will again meet at the starting point in 2772 i.e.,46 min and 12s.
According to the question,
1st number = 3M, 2nd number = 4M
where, M = HCF
But given, M = 4
We know that,
LCM = Product of two numbers / HCF
= (3M x 4M) / M
LCM = 12M = 12 x 4 = 48
Let the numbers be 66a and 66b, where a and b are co-primes.
According to the question,
66a + 66b = 1056
? 66(a + b) = 1056
? (a + b) = 1056/66 = 16
? Possible values of a and b are
(a = 1, b = 15), (a = 3, b = 13).
(a = 5, b = 11 ), (a = 7, b = 9)
? Numbers are
(66 x 1, 66 x 15), (66 x 3, 66 x 13),
(66 x 5, 66 x 11), (66 x 7, 66 x 9).
? Possible number of pairs = 4
Second number = (LCM x HCF) / first number
= (72 x 12)/ 24 = 36
Difference between the two numbers
= 36 - 24 = 12
LCM of 32, 40, 48 and 60 = 480
The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.
Hence, required number of numbers are 4.
HCF of (22n + 7, 33n + 10) is always 1.
lllustraction
For n = 1, HCF(29, 43) ? HCF = 1
For n = 2, HCF(51, 76) ? HCF = 1
For n = 3, HCF(73, 109) ? HCF = 1
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