JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 4)
A natural number has prime factorization given by n = 2x3y5z, where y and z are such
that y + z = 5 and y$$-$$1 + z$$-$$1 = $${5 \over 6}$$, y > z. Then the number of odd divisions of n, including 1, is :
that y + z = 5 and y$$-$$1 + z$$-$$1 = $${5 \over 6}$$, y > z. Then the number of odd divisions of n, including 1, is :
11
6
12
6x
Explanation
y + z = 5 ....... (1)
$${1 \over y} + {1 \over z} = {5 \over 6}$$
$$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$$
$$ \Rightarrow {5 \over {yz}} = {5 \over 6}$$
$$ \Rightarrow $$ yz = 6
Also, (y $$-$$ z)2 = (y + z)2 $$-$$ 4yz
$$ \Rightarrow $$ (y $$-$$ z)2 = (y + z)2 $$-$$ 4yz
$$ \Rightarrow $$ (y $$-$$ z)2 = 25 $$-$$ 4(6) = 1
$$ \Rightarrow $$ y $$-$$ z = 1 ..... (2)
from (1) and (2), y = 3 and z = 2
for calculating odd divisor of p = 2x . 3y . 5z
x must be zero
P = 20 . 33 . 52
$$ \Rightarrow $$ Total possible cases = (3050 + 3150 + 3250 + 3350 + .... + 3352)
$$ \therefore $$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12
$${1 \over y} + {1 \over z} = {5 \over 6}$$
$$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$$
$$ \Rightarrow {5 \over {yz}} = {5 \over 6}$$
$$ \Rightarrow $$ yz = 6
Also, (y $$-$$ z)2 = (y + z)2 $$-$$ 4yz
$$ \Rightarrow $$ (y $$-$$ z)2 = (y + z)2 $$-$$ 4yz
$$ \Rightarrow $$ (y $$-$$ z)2 = 25 $$-$$ 4(6) = 1
$$ \Rightarrow $$ y $$-$$ z = 1 ..... (2)
from (1) and (2), y = 3 and z = 2
for calculating odd divisor of p = 2x . 3y . 5z
x must be zero
P = 20 . 33 . 52
$$ \Rightarrow $$ Total possible cases = (3050 + 3150 + 3250 + 3350 + .... + 3352)
$$ \therefore $$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12
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