UKMT Special (Problem $14$)

No digit of the positive integer $N$ is prime. However, all the the single-digit primes divide $N$ exactly.

What is the smallest such integer $N$ ?

[UKMT Hamilton Olympiad $2016$ , H $2$ ]

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

You have until next Monday, $3:00$ pm!

Yajat Shamji - 6 months, 4 weeks ago

Step 1) The first statement implies that the possible digits of N are 0, 1, 4, 6, 8, and 9.

Step 2) The first statement implies that N is divisible by 2, 3, 5, and 7.

Step 3) Because 2 and 5 divide N exactly, 10 must also divide N. This implies that the last digit of N is 0.

Step 4) Now, lets try listing some positive integer that divisible by 3, 7, and 10.

   210, 420, 630, 840, ...

Notice that 840 is the smallest positive integer that satisfy the statement.

PS : I notice that the third step is not necessary, but it might give you insight on other problem.

Quest Keeper - 6 months, 3 weeks ago

Correct, but slightly different method.

Yajat Shamji - 6 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

UKMT Special (Problem \(14\))

Comments