NCERT Math Class 6th Question – 9 Exercise 3.7 |
Find the smallest 4-digit number which is divisible by 18, 24 and 32. Solution: The smallest 4-digit number = 1000.
To find the LCM of 18, 24 and 32, we have
∴ LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
Since, 288 is the smallest number which is exactly divisible by 18, 24 and 32.
But it is not a 4-digit number.
So, the multiple of 288 just above 1000 is: 1000 – 136 + 288 = 1152.
Hence, the required number is 1152. |
NCERT Math Class 6th Question – 1 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 2 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 3 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 4 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 5 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 6 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 7 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 8 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 10 Exercise 3.7 | Click here |
NCERT Math Class 6th Question – 11 Exercise 3.7 | Click here |
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