Find this integer with these characteristics

The positive integer is $15317$

The number ends in $17,$ so we have a starting point, the digit sum is $8.$ We now need a digit sum of 9, and the number appended on the left has to be divisible by $9$ . Since $17$ and $9$ are coprime, $17 \times 9 = 153$ is the smallest such number.

Hence the answer is $\boxed{15317}$

Source

Suppose n is the number then:

n ≡ 17(mod 100) since n ends in 17. n ≡ 17 (mod 9) since the digits of n sum to 17 n ≡ 17 (mod 17) since n is divisible by 17.

From Chinese Remainder Theorem,

n ≡ 17 mod 100⋅9⋅17 = 15300

and smallest positive solution n = 17+15300 = 15317

If the number $n$ ends in 17 and is divisible by 17, that means it's of the form $100m+17$ for some m a multiple of 17; the digit sum of m must be $17-1-7 = 9$ , implying that 9 divides m. Since 17 and 9 are coprime, the smallest possible value is $m = 17*9 = 153$ and $n = \boxed{15317}$ .

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# -*- coding: utf-8 -*-
"""
Created on Tue Nov 21 19:05:54 2017

@author: Michael Fitzgerald
"""

def enter_int():
    try:
        number = input("Enter number [10-99]: ")
        if (type(number) is int) and (len(str(number)) == 2): # checks whether entry is a number
            print "Valid entry."
            return number
    except:
       print 'Incorrect entry'

number = enter_int()

for i in range(number,1000000,100):
    #print i
    sum_ = 0
    nstr = str(i)
    for digit in nstr:
        sum_ += int(digit)
        #print sum_
    if (i%number == 0) and (sum_ == number):
        print '''
        The smallest positive integer which ends in %d, is divisible by %d, and whose digits sum to %d is %d
         ''' % (number, number, number, i)
        break

Enter number [10-99]: 17 Valid entry.

    The smallest positive integer which ends in 17, is divisible by 17, and whose digits sum to 17 is 15317

So the number where looking for ends in 17 which has an digit sum of 8 so we want the digits on the front to have a digit sum of 9 and be divisible by 17. We can see that the smaller possible integer is 17×9=153 (1+5+3=9) So the number is 15317

All those mathy approaches are... too mathy! Let's make things funnier by using a golfing language to solve this problem! I used Pyth , for instance:

f&q17sjT&!%T17q"17">`T_2

Which swiftly gave me the correct result, 15317.

How it works

f&q17sjT&!%T17q"17">`T_2 ~ Full program, outputs to STDOUT.

f                         ~ First input where the condition is truthy over [1, 2, 3, 4, ...]. 
  q17sjT;                 ~ Is (q) the sum (s) of the digits (jT;) 17?
 &                        ~ And
          !%T17           ~ Is the remainder when divided by 0 falsy (In other words, is T divisible by 17?) 
         &                ~ And 
               q"17">`T_2 ~ Are (q) the last two digits (>`T_2) equal to 17 ("17")?

Try it online!

let, the number is = x17 ; where x is unknown & sum of some digits & x+1+7= 17; hence, x = 9; & x is must be divisible by 17; now, we find 153/17= 9; so, x = 153; because, 1+5+3=9; hence, the number is x17= 15317; :)

End with 17 means 17 x #01, and the # is 17-1-7=9. Therefore 17x901 = 15317

The lazyiest solution is to use excel It always works fine unless you establish a successful model