Mathematics
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OpenStudy (anonymous):
Find the LCM of the largest 2-digit integer, the square of eleven and the square of nine.
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OpenStudy (anonymous):
@kropot72
OpenStudy (anonymous):
@jim_thompson5910 ,@Compassionate ,@tkhunny
ganeshie8 (ganeshie8):
whats the largest 2-digit integer ?
OpenStudy (anonymous):
99*99=9801
OpenStudy (anonymous):
squARE Of 11 is132 and 9 is 81
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ganeshie8 (ganeshie8):
thats right ! i would have worked it by prime factorization..
OpenStudy (anonymous):
lol square of 11 is 121
OpenStudy (anonymous):
ya @Yahoo!
OpenStudy (anonymous):
so answer is 9801
ganeshie8 (ganeshie8):
99 = 9.11 = 3^2.11
11^2 = 11^2
9^2 = 3^4
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OpenStudy (anonymous):
121*81=9801
ganeshie8 (ganeshie8):
LCM = 3^4*11^2 = 9801
OpenStudy (anonymous):
How many numbers divisible by each of the numbers 21, 36 and 66 are there, such that they are less than 10, 000?
ganeshie8 (ganeshie8):
the least number thats divisible by given set of numbers is their LCM
OpenStudy (anonymous):
2772
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ganeshie8 (ganeshie8):
thats the LCM
ganeshie8 (ganeshie8):
so any number of form 2772k will divisible all 3 numbers 21,36 and 66
ganeshie8 (ganeshie8):
how many numbers are there of that form under 10,000 ?
OpenStudy (anonymous):
9999/2772 =remainder 1683
9999-1683=ans?
ganeshie8 (ganeshie8):
nope, you need to find the multiples of 2772 :
2772x1 = 2772
2772x2 = ?
2772x3 = ?
2772x4 = ?
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ganeshie8 (ganeshie8):
stop when you exceed 10,000
OpenStudy (anonymous):
its upto 3
OpenStudy (anonymous):
8316
ganeshie8 (ganeshie8):
yes, there are exactly 3 numbers less than 10,000 that will be divisible by all 21,36 and 66