Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

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1

find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

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Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

Answer: To find the smallest number that satisfies these conditions, we need to set up a system of congruences based on the given remainders:

  1. The number leaves a remainder of 4 when divided by 5.
  2. The number leaves a remainder of 5 when divided by 6.
  3. The number leaves a remainder of 6 when divided by 7.
  4. The number leaves a remainder of 7 when divided by 8.
  5. The number leaves a remainder of 8 when divided by 9.

We can represent these conditions using congruences:

  1. ( x \equiv 4 \ (\text{mod} \ 5) )
  2. ( x \equiv 5 \ (\text{mod} \ 6) )
  3. ( x \equiv 6 \ (\text{mod} \ 7) )
  4. ( x \equiv 7 \ (\text{mod} \ 8) )
  5. ( x \equiv 8 \ (\text{mod} \ 9) )

Notice a pattern in these congruences. Each congruence can be rewritten in the form:
[ x \equiv -1 \ (\text{mod} \ n) ]

For each modulus ( n ):
x + 1 \equiv 0 \ (\text{mod} \ n)

Thus, we can rewrite the system as:
[ x + 1 \equiv 0 \ (\text{mod} \ 5) ]
[ x + 1 \equiv 0 \ (\text{mod} \ 6) ]
[ x + 1 \equiv 0 \ (\text{mod} \ 7) ]
x + 1 \equiv 0 \ (\text{mod} \ 8)
x + 1 \equiv 0 \ (\text{mod} \ 9)

This implies:
x + 1 \equiv 0 \ (\text{mod} \ \text{LCM}(5, 6, 7, 8, 9))

To solve for ( x ), we need to find the least common multiple (LCM) of the moduli 5, 6, 7, 8, and 9.

Finding the LCM:

  • The prime factorizations are:

    • ( 5 = 5 )
    • ( 6 = 2 \times 3 )
    • ( 7 = 7 )
    • ( 8 = 2^3 )
    • ( 9 = 3^2 )

  • The LCM is found by taking the highest power of each prime that appears:

    • ( 2^3 ) (from 8)
    • ( 3^2 ) (from 9)
    • ( 5 ) (from 5)
    • ( 7 ) (from 7)

Thus, the LCM is:
\text{LCM} = 2^3 \times 3^2 \times 5 \times 7

Calculating the LCM:
2^3 = 8
3^2 = 9
8 \times 9 = 72
72 \times 5 = 360
360 \times 7 = 2520

So, the LCM is 2520.

Therefore, ( x + 1 \equiv 0 \ (\text{mod} \ 2520) ), which means:
[ x + 1 = 2520k ]

For the smallest positive ( x ), let ( k = 1 ):
[ x + 1 = 2520 ]
[ x = 2520 - 1 ]
[ x = 2519 ]

Thus, the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8, and 8 on division by 9 is \boxed{2519}.

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