Question

find the smallest 3 digit number which when increased by 5 and divisible by 48,32,and 24​

Answer 1

Answer:

L.C.M of 24,32,36,54

2∣24,32,36,54

_______________

2∣12,16,18,27

_______________

9∣6,8,9,27

_______________

2∣6,8,1,3

_______________

3∣3,4,1,3

_______________

4∣1,4,1,1

_______________

1,1,1,1

=2×2×2×3×4×9

=8×12×9

=96×9

=864

∴ Required number = L.C.M of (24,32,36 and 54) −5

=864−5=859

Step-by-step explanation:

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Answer 2

Answer:

187

Step-by-step explanation:

first take the LCM of 48,32,24

which is 96

but this is not a three digit number

so just add 96+96

192

so just subtract 5 from it

which 187

no. is 187.