I am posting this as a note because I do not know the exact answer for this...Do post the solutions if u can crack this one!!

A set of positive integers satisfies the property that when 10^{20} , 15^{10} and 24^{15} are divided by any number in this set, at least one of the remainders is zero. What is the total number of elements in this set? a)1256 b)1266 c)1024 d)none of these

#NumberTheory

Note by Manu Mehta
6 years, 9 months ago

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Let A denote numbers divisible by 1020=22052010^{20}=2^{20}5^{20}, B denote numbers divisible by 1510=31051015^{10}=3^{10}5^{10} and C denote numbers divisible by 2415=24531524^{15}=2^{45}3^{15}.

What you want is a+b+c+d+e+f+ga+b+c+d+e+f+g.

g=Numbers divisible by all 3. So g has one number {1} so g=1\color{#3D99F6}{1}

d+g=Numbers in A and C= divisible by 2202^{20}. Hence d+g=21 or d=20\color{#3D99F6}{20}

f+g=Numbers in B and C= divisible by 3103^{10}. Hence f+g=11 or f=10\color{#3D99F6}{10}

e+g=Numbers in A and B= divisible by 5105^{10}. Hence e+g=11 or e=10\color{#3D99F6}{10}

a+d+e+g=Numbers in A=divisible by 102010^{20}. a+d+e+g=441 or a=410\color{#3D99F6}{410}

b+f+e+g=Numbers in B=divisible by 151015^{10}. b+f+e+g=121 or b=100\color{#3D99F6}{100}

c+d+f+g=Numbers in C=divisible by 241524^{15}. c+d+f+g=736 or c=705\color{#3D99F6}{705}

Adding, we get a+b+c+d+e+f+g=1256\color{#D61F06}{\boxed{1256}}

Pranjal Jain - 6 years, 6 months ago
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