The lowest number which leaves 4 as a remainder when divided by 10, 50 and 15 is

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Solution

Let's find LCM of $10, 50,$ and $15$ first.

LCM $= 2 \times 3 \times 5 \times 5 = 150$
Clearly, $150$ is completely divisible by $10, 50,$ and $15$ .
But, we want the remainder to be $4$ . So, we will add $4$ to $150$ .
i.e $150 + 4 = 154$ .
$∴ 154$ is the required number.

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