What two-digit positive integer is three less than a multiple of 7 and three less than a multiple of 8? - AIKO, infinite ways to autonomy.

Understanding the Context

We are searching for a two-digit integer x such that:
x ≡ –3 (mod 7) and x ≡ –3 (mod 8)
This is equivalent to:
x + 3 ≡ 0 (mod 7) and x + 3 ≡ 0 (mod 8)
Thus, x + 3 is a common multiple of 7 and 8.

Since 7 and 8 are coprime, their least common multiple is:
LCM(7, 8) = 56
So x + 3 = 56k, for some positive integer k

Then:
x = 56k – 3

Now, we seek two-digit values of x, so:
10 ≤ 56k – 3 ≤ 99
Add 3 to all parts:
13 ≤ 56k ≤ 102
Divide by 56:
13/56 ≤ k ≤ 102/56
0.23 ≤ k ≤ 1.82

Key Insights

Only integer value of k in this range is k = 1

Then:
x = 56 × 1 – 3 = 53

Check:

• Is 53 three less than a multiple of 7?
  56 is a multiple of 7, and 56 – 3 = 53 → ✅
  
• Is 53 three less than a multiple of 8?
  64 is a multiple of 8, and 64 – 3 = 61 → Wait! That doesn’t match.

Hold on—53 + 3 = 56, and 56 is a multiple of both 7 and 8.
So 53 ≡ –3 mod 7 and 53 ≡ –3 mod 8 — that’s correct.
But are 56 divisible by 8? Yes:
56 ÷ 8 = 7, so 56 is a multiple of 8 → ✅

Wait — but 56 – 3 = 53, and:

• 53 + 3 = 56 = 8 × 7 → multiple of 8 ✅
  
• 53 + 3 = 56 = 7 × 8 → multiple of 7 ✅

Final Thoughts

So 53 is three less than a common multiple of 7 and 8.
But is 56 the only multiple of 56 in the two-digit range?
Next multiple: k = 2 → x + 3 = 112 → x = 109, which is three digits → too large.

Only valid value is k = 1 → x = 53

Double-check:

• 53 + 3 = 56
  
• 56 ÷ 7 = 8 → divisible ✅
  
• 56 ÷ 8 = 7 → divisible ✅

Thus, 53 satisfies both conditions: it is three less than 56, a multiple of both 7 and 8.

🔍 Conclusion: The two-digit positive integer that is three less than a multiple of both 7 and 8 is 53.