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How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?

I am able to get the seven-digit numbers divisible by 11

and

I am also able to get the seven-digit numbers whose sum of their digits equal to 59.

But i am not able to get how i can get the count of 7 digit numbers satisying both the condition.

Thanks in advance.

Thanks in advance.

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Since I promised vikiiii that I’d answer, here’s my version.

A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. Say that the seven-digit number is $abcdefg$, where the letters represent the individual digits; then it’s divisible by $11$ if and only if $(a+c+e+g)-(b+d+f)$ is a multiple of $11$. Let $S=a+c+e+g$ and $T=b+d+f$; then we need $S-T$ to be a multiple of $11$ and $S+T=59$.

Since $S+T=59$, one of $S$ and $T$ must be odd and the other even, so their difference must be odd. Thus, $S-T$ cannot be $0$ or $\pm22$. We should look for ways to make it $\pm11$ or $\pm33$. (It clearly can’t be any bigger in magnitude, since $S\le 4\cdot9=36$.)

Suppose that $S-T=11$; then $70=11+59=(S-T)+(S+T)=2S$, so $S=35$ and $T=24$. This is possible only if three of the digits $a,c,e,g$ are $9$ and the fourth is $8$; there’s no other way to get four digits that total $35$. For three digits to total $24$, they must average $8$, so the only possibilities are that all three are $8$, that two are $9$ and one is $6$, or that they are $7,8$, and $9$. Thus, the digits $a,c,e,g$ in that order must be $8999,9899,9989$, or $9998$, and the digits $b,d,f$ must be $888,699,969,996,789,798,879,897,978$, or $987$, for a total of $4\cdot 10=40$ numbers.

Now suppose that $S-T=-11$; then by similar reasoning $2S=-11+59=48$, so $S=24$, and $T=35$. But $T\le 3\cdot9=27$, so this is impossible. Similarly, $S-T$ cannot be $-33$. The only remaining case is $S-T=33$. Then $2S=33+59=92$, and $S=46$, which is again impossible. Thus, the first case contained all of the actual solutions, and there are $40$ of them.

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  • $\begingroup$ M.Scott you are genius.Thanks so much.This is the best way.Thank you again. $\endgroup$ – vikiiii Mar 30 '12 at 15:13
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If we look at the number $a_6a_5a_4a_3a_2a_1a_0$: $$a_6a_5a_4a_3a_2a_1a_0 = a_610^6+a_510^5+ \dots +a_010^0 \\ \equiv a_6(-1)^6 + \dots+a_0(-1)^0 \mod 11 = a_6 -a_5+a_4-a_3+a_2-a_1+a_0 \mod 11$$ So if the number will be divisible by 11 we require $$a_6 -a_5+a_4-a_3+a_2-a_1+a_0 = 11m$$

We take note that $0 \leq a_i \leq 9$ so possible values of $11m$ are limited to: $$-27 = 4\cdot0-3\cdot9\leq 11m \leq 4\cdot 9-3\cdot0 = 36$$ Which really means that $11m \in \{-22,-11,0,11,22,33\}$

The other requirement of the question was that $a_6 + \dots +a_0 = 59$. From this and the above equation (not sure how to number them and align them nicely in TeX) we add and subtract and get much nicer equations:

$$a_6 +a_4+a_2+a_0 = \frac{59+11m} 2 \\a_5+a_3+a_1 =\frac{ 59-11m}2$$

Of course the LHS is whole, so the right hand side must be as well, which means $m$ needs to be odd. So we reduce our options to:

$$11m \in \{-11,11,33\} \implies \frac{59+11m} 2 \in \{24,35,46\}$$ Of course the sum of four digits can't be $46$ from our above inequality on $a_i$, and similarly $$\frac{59-11m} 2 \in \{35,24\}$$ But the sum of three digits can't be $35$, so we're left with $$a_6 +a_4+a_2+a_0 = 35 \\a_5+a_3+a_1 =24$$

It's easy to see that the only options for the four digits is a permutation of $9998$, and the three digits must be a permutation of one of $\{699,789,888\}$.

Order doesn't matter, so basic combinatorics gives $4\cdot(3 + 3! + 1) = 40$ such numbers.

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  • $\begingroup$ Nice solution davin $\endgroup$ – juantheron Nov 8 '13 at 13:12

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