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(a) __ 6724

(b) 4765 __ 2

Answer :

(a) ___ 6724

Sum of the digits = 4 + 2 + 7 + 6 = 19

The smallest digit to be placed is blank space = 2

Then the sum = 19 + 2 = 21 which is divisible by 3.

The greatest digit to be placed in blank space = 8

Then, the sum = 19 + 8 = 27 which is divisible by 3

Hence, the required digits are 2 and 8.

(b) 4765 ____ 2.

Sum of digits = 2 + 5 + 6 + 7 + 4 = 24

The smallest digits to be place in blank space = 0

Then, sum = 24 + 0 = 24

which is divisible by 3.

The greatest digit to be placed in blank space = 9.

Then, the sum = 24 + 9 = 33 which is divisible by 3.

Hence, the required digits are 0 and 9.

- Q.1 Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes or no):
- Q.2 Using divisibility tests, determine which of the following numbers are divisible by 4; by 8: (a) 572 (b) 726352 (c) 5500 (d) 6000 (e) 12159 (f) 14560 (g) 21084 (h) 31795072 (i) 1700 (j) 2150
- Q.3 Using divisibility tests, determine which of following numbers are divisible by 6: (a) 297144 (b) 1258 (c) 4335 (d) 61233 (e) 901352 (f) 438750 (g) 1790184 (h) 12583 (i) 639210 (j) 17852
- 4. Using divisibility tests, determine which of the following numbers are divisible by 11: (a) 5445 (b) 10824 (c) 7138965 (d) 70169308 (e) 10000001 (f) 901153
- Q.6. Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11: (a) 92 __ 389 (b) 8 __ 9484

- NCERT solutions for class 6 maths chapter 1 Knowing Our Numbers
- NCERT solutions for class 6 maths chapter 2 Whole Numbers
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- NCERT solutions for class 6 maths chapter 12 Ratio and Proportion
- NCERT solutions for class 6 maths chapter 13 Symmetry
- NCERT solutions for class 6 maths chapter 14 Practical Geometry

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