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When 12, 16, 18, 20 and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousand's place in x?
Answer & Solution
Correct Answer:
Option
D
LCM of 12, 16, 18, 20, 25 = 3600
For remainder 4 = 3600k + 4
If divisible by 7 $$ = \frac{{7 \times 514k + 2k + 4}}{7}$$
$$\frac{{2k + 4}}{7},$$ If k = 5 divisible
x = 3600 × 5 + 4
= 18000 + 4
$$ = \mathop {\mathop {18004}\limits_ \downarrow }\limits_{\boxed8} $$
For remainder 4 = 3600k + 4
If divisible by 7 $$ = \frac{{7 \times 514k + 2k + 4}}{7}$$
$$\frac{{2k + 4}}{7},$$ If k = 5 divisible
x = 3600 × 5 + 4
= 18000 + 4
$$ = \mathop {\mathop {18004}\limits_ \downarrow }\limits_{\boxed8} $$
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