Question: What two-digit positive integer is one more than a multiple of 7 and three more than a multiple of 5? - RTA

Understanding the Context

Mathematical reasoning puzzles often emerge during periods of heightened digital engagement. With the rise of short-form content and micro-learning on platforms like discover, questions like this tap into a collective desire for clarity, precision, and repeatable patterns. Younger and adult audiences alike appreciate challenges that blend logic with real-world relevance—especially when framed as answers to “How do you know?” rather than abstract riddles. This query combines fundamental number theory with practical thinking, making it resonant for users curious about cognitive patterns, automation logic, and structured inference—traits increasingly relevant in modern life.

How Does This Question Actually Work?

To solve What two-digit positive integer is one more than a multiple of 7 and three more than a multiple of 5? start with the clues:

• The number is one more than a multiple of 7 → it leaves a remainder of 1 when divided by 7.
• It’s three more than a multiple of 5 → when divided by 5, it leaves a remainder of 3.

Translating this into math:
We’re looking for a two-digit number x such that:

• x ≡ 1 (mod 7)
• x ≡ 3 (mod 5)

Key Insights

Using the Chinese Remainder Theorem or step-by-step testing, test combinations that satisfy both conditions. From 10 to 99, only in two candidates meet both: 53 and a secondary number outside two digits. Among two-digit integers, 53 stands out:

• 53 ÷ 7 = 7×7 + 4 → but wait—actually 7×7=49 → 49 + 1 = 50 → correction: 53 = 7×7 + 4? Let’s double-check: 7×7 = 49 → 49 + 1 = 50, so 53 is not 1 more than 7×k. Re-evaluating:
  Try values 1 more than multiples of 7:
  8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99
  Now check which leave remainder 3 when divided by 5:
  53 ÷ 5 = 10×5 + 3 → fits
  Check 50: 50 ÷ 5 = 10, remainder 0 → no
  Try 58: not in list
  71: 71 ÷ 5 = 14×5 + 1 → no
  Only 53 satisfies both conditions when confirmed:
• 53 – 1 = 52 → is 52 divisible by 7? 7×7=49, 52–49=3 → no
  Wait—recheck: Problem says one more than multiple of 7 → x – 1 divisible by 7
  53 – 1 = 52 → 52 ÷ 7 = ~7.4 → not divisible. Mistake!

Correct logic:
Solve system:
x ≡ 1 (mod 7) → x = 7a + 1
x ≡ 3 (mod 5) → x = 5b + 3

Set them equal: 7a + 1 = 5b + 3 → 7a – 5b = 2
Try small integer values for a:
a = 1: 7 → 7b = 5 → no
a = 2: 14 → 14 – 3 = 11 → 5b = 11 → no
a = 3: 21 → 21 – 3 = 18 → b = 18/5 → no
a = 4: 28 → 28 – 3 = 25 → b = 5 → valid
So x = 7×4 + 1 = 29 → check: 29 ÷ 5 = 5×5 + 4 → remainder 4 → does NOT work. Contradiction?

Wait: 29 mod 5 = 4 → invalid. Recalculate:
7a + 1 ≡ 3 mod 5 → 7a ≡ 2 mod 5 → 7 ≡ 2 mod 5 → so:
2a ≡ 2 mod 5 → a ≡ 1 mod 5 → a = 1, 6, 11…
Try a = 1: x = 7×1 + 1 = 8 → 8 mod 5 = 3 → yes!
x = 8 → check both:
8 ÷ 7 = 1×7 + 1 → good
8 ÷ 5 = 1×5 + 3 → good
But 8 is not two-digit.

Next: a = 6 → 7×6 + 1 = 43
43 ÷ 5 = 8×5 + 3 → remainder 3 → valid
Two-digit, works!

Final Thoughts

So 43 satisfies both:

• 43 – 1 = 42 → 42 ÷ 7 = 6 → whole
• 43 ÷ 5 = 8×5 + 3 → remainder 3

Now check for others: a = 11 → 7×11 + 1 = 78 → 78 ÷ 5 = 15×5 + 3 → remainder 3 → valid, two-digit again
x = 78 → also satisfies
But is there a unique two-digit answer? Let’s list all two-digit x = 7a + 1, a from 2 to 14 (since 7×14 + 1 = 99):
a = 2 → 15 → 15 mod 5 = 0 → no
a = 3 → 22 → 22 mod 5 = 2 → no
a = 4 → 29 → 29 mod 5 = 4 → no
a = 5 → 36 → 1 → no
a = 6 → 43 → 3 → yes
a = 7 → 50 → 0 → no
a = 8 → 57 → 2 → no
a = 9 → 64 → 4 → no
a = 10 → 71 → 1 → no
a = 11 → 78 → 3 → yes
a = 12 → 85 → 0 → no
a = 13 → 92 → 2 → no
a = 14 → 99 → 4 → no

Valid two-digit solutions: 43, 78

But 78 is two-digit, so both qualify? Wait—but the question asks for the two-digit integer—implying uniqueness. Recheck constraints.

Ah—problem likely assumes smallest, but both 43 and 78 are valid. However, in typical educational contexts, the focus is on 43 as the primary example, especially when phrased as “the” integer. But mathematically, 43 and 78 are both solutions.

However, the real intrigue lies in the dual logic—how two different numbers satisfy the equation. This nuance reflects real-world pattern recognition, where constraints can admit multiple valid outputs, prompting deeper exploration rather than a single “correct” answer.