求解满足n+m、n-m均被7整除的三位数n与其反转数m的数对

Let's break down this problem step by step to find all valid (n,m) pairs.

First, let's define our variables properly: let ( n = 100a + 10b + c ), where ( a ) is from 1 to 9 (since n is a three-digit number), and ( b,c ) are from 0 to9. The reversed number ( m = 100c +10b +a ). We need ( m \neq n ), and both ( n+m ) and ( n-m ) must be divisible by7.

Step 1: Use the divisibility of ( n - m ) #

Calculating ( n - m ):
[ n - m = (100a+10b+c) - (100c+10b+a) =99(a - c) ]
Since 99 is 9×11, neither of which are divisible by7, the only way 7 divides ( n-m ) is if 7 divides ( a - c ).

Given that ( a ) and ( c ) are digits, the only possible values for ( a - c ) are 7 or -7 (any larger multiple would require digits outside 0-9). This gives us two main cases:

• Case 1: ( a = c +7 ) → possible (a,c) pairs: (7,0), (8,1), (9,2)
• Case 2: ( c = a +7 ) → possible (a,c) pairs: (1,8), (2,9)

Step2: Use the divisibility of ( n + m ) #

Now let's look at ( n + m ):
[ n + m = (100a+10b+c)+(100c+10b+a)=101(a+c)+20b ]
To check divisibility by7, let's compute each term modulo7:

• 101 divided by7 is 14 with remainder3 → (101 \equiv3 \mod7)
  -20 divided by7 is2 with remainder6 → (20 \equiv6 \mod7)

So substituting into the equation:
[3(a+c) +6b \equiv0 \mod7]
We can divide both sides by3 (since 3 and7 are coprime, this doesn't change the divisibility):
[ (a+c) +2b \equiv0 \mod7 ]

Now let's work through each case to find valid b values.

Case1: (a = c +7)

• Subcase 1a: (a,c)=(7,0)
  Here (a+c=7). Plugging into our equation: (7 +2b \equiv0 \mod7). This simplifies to (2b \equiv0 \mod7), which means b must be a multiple of7. Valid digits for b are 0 and7.
  This gives us two pairs: (700,7) and (770,77). Both satisfy all conditions (m≠n, sums/differences divisible by7).

• Subcase1b: (a,c)=(8,1)
  Here (a+c=9). The equation becomes (9 +2b \equiv0 \mod7). 9 mod7 is2, so (2 +2b \equiv0 \mod7) → (b+1 \equiv0 \mod7). The only digit solution is b=6.
  This gives the pair (861,168). Checking: 861+168=1029 (divisible by7: 1029/7=147), 861-168=693 (divisible by7:693/7=99). Perfect.

• Subcase1c: (a,c)=(9,2)
  Here (a+c=11). 11 mod7 is4, so equation becomes (4 +2b \equiv0 \mod7) → (2b \equiv3 \mod7). Multiply both sides by4 (the inverse of2 mod7, since 2×4=8≡1 mod7): (b \equiv12 \equiv5 mod7). So b=5.
  This gives (952,259). Checking sums/differences: 952+259=1211 (1211/7=173), 952-259=693 (693/7=99). Valid.

Case2: (c = a +7)

• Subcase2a: (a,c)=(1,8)
  Same as Subcase1b: (a+c=9), so b=6. This gives (168,861), which is the reverse of the pair we found earlier. Valid.

• Subcase2b: (a,c)=(2,9)
  Same as Subcase1c: (a+c=11), so b=5. This gives (259,952), reverse of the earlier pair. Valid.

All Valid Pairs #

Putting it all together, the valid (n,m) pairs (with n being a three-digit number) are:

• (700,7)
• (770,77)
• (168,861)
• (861,168)
• (259,952)
• (952,259)

If the problem expects m to also be a three-digit number (i.e., no leading zeros in the reversed number), then we exclude the first two pairs, leaving only the four three-digit pairs.

内容的提问来源于stack exchange，提问作者sdfghjgfdscfvght