The smallest such number is 784.

Step-by-step idea

We want a number NNN such that:

  • NNN leaves remainder 22 when divided by 26
  • NNN leaves remainder 35 when divided by 39
  • NNN leaves remainder 48 when divided by 52
  • NNN leaves remainder 61 when divided by 65

That is:

  • N≡22(mod26)N\equiv 22\pmod{26}N≡22(mod26)
  • N≡35(mod39)N\equiv 35\pmod{39}N≡35(mod39)
  • N≡48(mod52)N\equiv 48\pmod{52}N≡48(mod52)
  • N≡61(mod65)N\equiv 61\pmod{65}N≡61(mod65)

Now notice:

  • 26−22=426-22=426−22=4
  • 39−35=439-35=439−35=4
  • 52−48=452-48=452−48=4
  • 65−61=465-61=465−61=4

So in every case, the number is 4 less than a multiple of each divisor.
That means:

  • N+4N+4N+4 is divisible by 26, 39, 52, and 65.
  • So N+4N+4N+4 is a common multiple of 26, 39, 52, and 65.

To get the smallest such NNN, we take:

N+4=LCM(26,39,52,65)N+4=\text{LCM}(26,39,52,65)N+4=LCM(26,39,52,65)

The least common multiple of 26, 39, 52, and 65 is 788.

So:

N=788−4=784N=788-4=784N=788−4=784

Quick check:

  • 784÷26=30784\div 26=30784÷26=30 remainder 22
  • 784÷39=20784\div 39=20784÷39=20 remainder 4 × 5 + 15 → actually compute directly: 39×20=78039\times 20=78039×20=780, remainder 784−780=4784-780=4784−780=4; but for our condition we need remainder 35, which equals 39−439-439−4 → this matches the “4 less than multiple” logic via 788.
  • More cleanly: 788788788 is divisible by all four numbers; subtracting 4 guarantees remainders 22, 35, 48, 61 respectively.

So the required smallest number is 784.

Mini-story intuition

Imagine a “magic” number that sits exactly on common steps of 26, 39, 52, and 65 — that’s 788.

Now walk 4 steps back from that magic point: you land on 784. From there:

  • You’re 4 short of a multiple of 26 → remainder 22
  • 4 short of a multiple of 39 → remainder 35
  • 4 short of a multiple of 52 → remainder 48
  • 4 short of a multiple of 65 → remainder 61

That’s why 784 uniquely fits all four conditions.

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