Divisible by 7

How many four digit positive integers divisible by 7 have the property that, when the first and last digits are interchanged, the result is a (not necessarily four-digit) positive integer divisible by 7?
Source: NCTM Mathematics Teacher, October 2006

Solution
Is 8673 divisible by 7? One way to find out is by determining if 8673 “behaves” like 7 or 14 or some other small multiples of 7. Modulo 7 arithmetic will help us do that. In its most basic definition, modulo arithmetic is the arithmetic of remainder. We say “8 is congruent to 15 modulo 7” and write 8\equiv 15\bmod 7, because 8 and 15 yield the same remainder when divided by 7.
Using 8673=8(1000)+6(100)+7(10)+3(1), we will do the same calculation but with congruent numbers mod 7. For ease in presentation we drop the modulo 7 notation.
Since 1000\equiv 6,100\equiv 2,10\equiv 3,8\equiv 1,7\equiv 0,
8(1000)+6(100)+7(10)+3(1)\equiv 1(6)+6(2)+0(3)+3(1)
\equiv 6+12+0+3\equiv 21
8673 is divisible by 7.

Let axy\,b be a 4-digit positive integer where the digits are a,x,y, and b. Suppose 7 divides axy\,b and bxya. Show that 7 divides (a-b).
Since 7 divides axy\,b and bxya, there exist integers k_1 and k_2 such that
a(1000)+x(100)+y(10)+b=7k_1\qquad\qquad (1)
b(1000)+x(100)+y(10)+a=7k_2\qquad\qquad (2)
Subtract Eq. (2) from Eq. (1)
(a-b)1000+(b-a)=7k_3 where k_3=k_1-k_2
1000a-1000b+b-a=7k_3
999(a-b)=7k_3
In other words, 7 divides 999(a-b). Since 7 does not divide 999, 7 must divide (a-b).

Show that if 7 divides axy\,b, then 7 divides 10x+y.
Given a(1000)+x(100)+y(10)+b=7k_4 for some integer k_4,
1000a+100x+10y+b=7k_4
1001a-a+100x+10y+b=7k_4
1001a-(a-b)+100x+10y=7k_4
Since 7 divides 1001a and (a-b), 7 must divide 100x+10y. Hence, 7 divides 10x+y.

We have two cases to consider: a=b and a\neq b.
Case 1: a=b
The integers are of the form axy\,a where the xy numbers are divisible by 7. The 15 possible xy numbers are: 00,07,14,21,28,35,42,49,56,63,70,77,84,91,98. The 9 possible a digits are: 1,2,3,4,5,6,7,8,9. There are 9\times 15=135 possible axy\,a.
Cases 2: a\neq b
The integers are of the form axy\,b where digits a and b are congruent. Interchanging congruent digits like 0 and 7, 1 and 8, 2 and 9 has no bearing on the remainder. The 5 possible axy\,b are: 1xy8,8xy1,2xy9,9xy2,7xy0. There are 5\times 15=75 possible axy\,b.
Total number of integers equals 135+75=210.
List of some of the integers
1001,1008,1071,1078,1141,1148,1211,1218,1281,1288,1351,1358,1421,1428,1491,1498
\cdots
7000,7007,7070,7077,7140,7147,7210,7217,7280,7287,7350,7357,7420,7427,7490,7497
\cdots
9002,9009,9072,9079,9142,9149,9212,9219,9282,9289,9352,9359,9422,9429,9492,9499

Answer: 210

 

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , , . Bookmark the permalink.

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