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The sum of the two smallest positive divisors of an integer $N$ is $6$, while the sum of the two largest positive divisors of $N$ is $1122$. Find $N$.

I came across this question in a Math Olympiad Competition. I am able to find out that the two smallest positive divisors would be $1$ and $5$ but after that, I am not sure how to work on to find out the value of $N$. Can anyone help? Thanks.

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4 Answers 4

up vote 15 down vote accepted

Hint: The two largest positive divisors of $N$ are $N$ and $N/5$.

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Could you please share why the two largest positive divisors of $N$ are $N$ and $N/5$ Why not $N/3$. Thanks. –  Gaurang Tandon Jun 4 at 14:32
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@GaurangTandon, if $N/d$ is a divisor of $N$ then $d$ is a divisor of $N$. This means that $d\ge 5$, in this case. –  lhf Jun 4 at 14:35
    
Oh I see. Thanks :) –  Gaurang Tandon Jun 4 at 14:42
    
@Gaurang See my answer, where I explain the essence of the matter. –  Bill Dubuque Jun 4 at 16:00

Thanks to the hints from lhf, I am now able to solve:

Smallest Positive Divisors: 1 and 5 Largest Positive Divisors: $N$ and $N/5$.

Basically: $N$ $+$ $N/5$ = 1122

$6N$ = 5610

$N$ = 935

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Hint $ $ The set of factors of $\,N\,$ enjoy a cofactor involution (reflection) $\ k\mapsto k' = N/k,\,$ so $\,N = k\,k'.\,$ This gives a pairing of $\,k\,$ with its cofactor $\,k' = N/k.\,$ It's order reversing $\, j < k\,\Rightarrow\, N/j > N/k,\,$ so it pairs the least factor $\,k_1$ with the greatest $\,N/k_1;\,$ 2nd least $\,k_2$ with the 2nd greatest $\,N/k_2,\,$ etc.

Thus the least factor $\,k_1 = 1$ pairs with the greatest $\,N/1 = N.\ $ The 2nd least factor $\,k_2$ is the least prime factor $\,p\,$ (else some prime $\,q\mid k_2\mid n,\,$ and $\,q\le k_2< p,\,$ contra leastness of $\,p).$ Therefore this 2nd least factor $\,k_2 =p\,$ pairs with $\,N/p =\,$ 2nd greatest factor. The rest is straightforward.

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Let $$ N=p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}, p_i>p_{i+1} $$ be the canonical decomposition of the integer number $N.$ The first two smallest divisors are $1$ and $p_1$. The largest two divisors are $p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}$ and $p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}$. Thus we obtain the following system $$ 1+p_1=6\\ p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}+p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}=1122 $$ From the fist equation we obtain that $p_1=5.$ Rewrite the second one $$ p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}+p_1^{k_1}p_2^{k_2}\cdots p_n^{k_n}=p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}(1+p_1)=6p_1^{k_1-1}p_2^{k_2}\cdots p_n^{k_n}=1122. $$ At last $$ N=p_1 \frac{1122}{6}=935. $$

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