Question

find the least number of four digits which exactly divisible by 32,36,40,45,48​

Answer 1

Answer:

100

Step-by-step explanation:

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.

80The 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]

Answer 2

Step-by-step explanation:

we find the LCM of 32,36,40,45,48

32= 2×2×2×2×2

36 = 2×2×3×3

40 =2×2×2×5

L.C.M = 288

The smallest four digit no is 288×4

1152

this number is divisible by all digit 32,36,40,45,48