Find the least number which when divided by 12, 16, 18, 21 and 28 leaves the same remainder 3 in each case
Answers
First of all find LCM and as it leaves remainder of 4 then the number is LCM +4
12=2×2×3=2^2×3
16=2×2×2×2=2^4
18=2×3×3=3^2×2
30=2×3×5
LCM (12,16,18,30)=2^4×3^2×5
LCM=16×9×5
LCM=720
The number can be 720+4=724
But the condition is, the number should also be completely divisible by 7
As 724 is not completely divisible by 7
Therefore the number is multiple of 720 plus 4 which is divisible by 7
720×2+4=1444 not divisible by 7
720×3+4=2164 not divisible by 7
720×4+4=2884
2884 is completely divisible by 7
Answer:
the answer is
Step-by-step explanation:
Take the LCM of 9, 12, 16 and 30.
LCM ( 9, 12, 16, 30 ) = 720.
720 is the least no. that can divide these no. leaving remainder 0.
We have to find out the number that divides leaving remainder 3 in each case.
So, we add 3 to 720.
➡ 720 + 3 = 723.
Required no is 723.
Let's check this number.
➡ 723 ÷ 9 => q = 80 and r = 3
➡ 723 ÷ 12 => q = 60 and r = 3
➡ 723 ÷ 16 => q = 45 and r = 3
➡ 723 ÷ 30 => q = 24 and = r = 3
Hope it will help u !!