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Number Theory Level 2

What is the smallest four digit multiple of 17, the sum of its digit should be 17.

The answer is 1088.

3 solutions

Naren Bhandari
May 24, 2018

Divisibility rule of 17

Subtract 5 times the last of digit from the remaining truncated number. If the result is divisible by 17 then the original number is also divisible by 17.

Call $n =\overline{abcd}$ the smallest four digit number. As per the divisibility test of $17$ . We have new number $m =\overline{abc} -5d =10^2a+10b +c -5d$ . Assume number is divisible by $17$ . Then we get $17 m =10^2a+10b+c-5d = 99a + 9b -6d +(a+b+c+d) \\17(m-1)=3(33a+3b-2d)\implies 17 \left(\dfrac{m-1}{3} \right) = 33a+3b-2d$ Clearly, left hand side is $m\equiv 1 \mod(3)$ which implies $m = 3k+1$ for positive integers $k$ . For the smallest possible value of RHS the value of LHS must be as small as possible which is only If $m=4$ . Then $33a +3b -2d = 1\times 17 =17$ Notice $a\neq 0$ so the smallest possible value of $a=1$ then last equation becomes $33 + 3b-2d = 17 \implies 2d-3b =16\implies d =\dfrac{16+3b}{2}$ For $b = 0, 2$ then $d = 8, 11$ respectively. Since $d \leq 9, \therefore d =8 , b =0, a =1$ & $c = 17- 9 =8$ which makes $\boxed{1088}$ as smallest required number.

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Further $b = \dfrac{17\left(\dfrac{m-1}{3}\right) - 33a +2d}{3} < 30, \quad m>4$

For $m=7 , b = \dfrac{34-33a+2d}{3}$ , note that $a\neq 0 , a < 2 \therefore a=1$ . $b = \dfrac{1+2d}{3}$ Thus value of $d$ are $4,7$ and solution for $b = 3, 5$ Hence $\overline {abcd }= 1387 , 1547$

For $m = 10 , b = \dfrac{51 -33a + 2d}{3}$ the possible value of $a$ are $1$ and $2$ . For $a = 1, b = \dfrac{18 +2d}{3}< 30$ . For $d = 0 , b=6$ but $c =10$ so this is rejected . Now for $d =3, b = 8, c = 7 \therefore \overline{abcd} = 1873$ . For $a =2, b = \dfrac{2d -15}{3}$ . Only possible value of $d =9$ for which $b = 1, c = 5$ . Therefore, $\overline{abcd} = 2159$ .

For $m = 13 , b = \dfrac{68 -33a + 2d}{3}$ . Only satisfying value of $a =2$ which results $d =5 ,8, b = 4, 6 , c = 6,1 \implies \overline{abcd} = 2465, 2618$ .

On the similar manner other such number can also be obtained.

Naren Bhandari - 3 years ago

Giorgos K.
May 15, 2018

$Mathematica$

Min@Select[17 Range[Ceiling[1000/17],Floor[9999/17]],Total@IntegerDigits@#==17&]

$\boxed{1088}$

Abhay Singh
May 15, 2018

We know that that a number, which is divisible by 9,the seed (sum of the digits till a single digit is obtained) of that number must be 9. This also implies that when a multiple of 9 is added to some other number, the seed of that (some)number will remain unchanged. So, we can write the number as N = 17×9×n +17. The first 17 is to get a multiple of 17, 9 is to make seed = 9, whereas n is to get the smallest possible four digit number. Also, last 17 will play the role of making N's seed = 8, which gives a possibility that N's digit sum will be 8, which may give sum of the digits as 17. Put n = 7 to get the required answer.

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The answer is 1088.

3 solutions

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