the question is find the least no of 5 digits which is exactly divisible by 32,36,40,45and 48
it's ok rose I also didn't get the question this is the question of sixth class
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the question is find the least no of 5 digits which is exactly divisible by 32,36,40,45and 48
it's ok rose I also didn't get the question this is the question of sixth class
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Answer:
A number which is divisible by 32,36,40,45,48 must also be divisible my their LCM.
So, we find the LCM of 32,36,40,45,48
\begin{gathered}32 = 2 {}^{5} \\ 36 = 2 {}^{2} \times 3 { }^{2} \\ 40 = 2 {}^{3} \times 5 \\ 45 = 3 {}^{2} \times 5 \\ 48 = 2 {}^{4} \times 3\end{gathered}
32=2
5
36=2
2
×3
2
40=2
3
×5
45=3
2
×5
48=2
4
×3
so \: LCM \: = 2 {}^{5} \times 3 { }^{2} \times 5 = 1440soLCM=2
5
×3
2
×5=1440
=>Now we need to find the smallest five digit number divisible by 1440.
The smallest five digit number is 10000.
10000 = 6×1440 +1360
That is, on dividing 10000 by 1440 quotient is 6
So, (7×1440) is the smallest required five-digit number
ANSWER: [ 7×1440 = 10080]
Step-by-step explanation:
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Answer:
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