Q1. Using the digits 4, 0 and 5, form all possible 3 digit numbers without repeating the digits. Also, classify them in a multiple of 2 and 3.
Solution
Digits are: 4, 0, 5
3 digit numbers are: 405, 450, 540, 504
A number is multiple of 2 if its unit place is even.
Therefore, 450, 540 and 504 are multiples of 2.
A number is multiple of 3 if the sum of digits is divisible by 3.
For all the above 3 digit numbers, sum of digits is 9, which is divisible by 3.
Hence all numbers are divisible by 3.
Q2. Write all possible numbers using the digits 2, 4, 7 without repeating them.
Solution
All possible numbers with digits 2, 4 and 7 are: 247, 274, 427, 472, 742, 724.
Q3. Write the following numbers in usual form: (i) 100 x 7 + 10 x 9 + 1 x 8 (ii) 1000 x 3 + 100 x 1 + 10 x 5 + 1 x 9
Solution
(i) 100 x 7 + 10 x 9 + 1 x 8 = 700 + 90 + 8 = 798 (ii) 1000 x 3 + 100 x 1 + 10 x 5 + 1 x 9 = 3000 + 100 + 50 + 9 = 3159
Q4. Test the divisibility of 46602479 by 11.
Solution
A number is divisible by 11 if the difference of the sum of the digits at even places and the sum of the digits at odd places is divisible by 11.
Consider 46602479.
Sum of digits at odd places = 4 + 6 + 2 + 7 = 19
Sum of digits at even places = 6 + 0 + 4 + 9 = 19
Difference = 19 - 19 = 0
Since, 0 is divisible by 11.
Hence, 46602479 is divisible by 11.
Q5. Which of the following numbers are divisible by 4?
(i) 45748 (ii) 21404
Solution
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
(i) Consider 45748
Number formed by last two digits is 48.
Since, 48 is divisible by 4.
Hence, 45748 is also divisible by 4.
(ii) Consider 21404
Number formed by last two digits is 04.
Since, 04 is divisible by 4.
Hence, 21404 is also divisible by 4.
Q6. "The difference of the two digit chosen number and the number obtained by reversing the digits is divisible by 9." Verify this property for the numbers 67 and 39.
Solution
Consider the number 67. Number after reversing the digits is 76. Difference = 76 - 67 = 9 Clearly the difference is divisible by 9. Again, consider the number 39. Number after reversing the digits is 93. Difference = 93 - 39 = 54 = 9 x 6 Clearly the difference is divisible by 9.
Q7. Test the divisibility of the following numbers by 3.
(i) 1956 (ii) 15693
Solution
(i) Given number 1956.
Sum of digits = 1 + 9 + 5 + 6 = 21
Since 21 is divisible by 3.
Therefore, 1956 is also divisible by 3.
(ii) Given number 15693.
Sum of digits = 1 + 5 + 6 + 9 + 3 = 24
Since 24 is divisible by 3.
Hence the number 15693 is divisible by 3.
Q8. Complete the series: (i) 20, 15, 10, ___, ___. (ii) 3, 9, 15, ___, ___.
Solution
(i) Here each term is obtained by subtracting 5 from previous term, therefore series is 20, 15, 10, 5, 0 (ii) Here, each term is an odd multiple of 3, therefore series is 3, 9, 15, 21, 27.
Q9. "The sum of a two digit number and the number obtained by reversing its digits is divisible by 11." Verify this property for the numbers 32 and 89.
Solution
Consider the number 32. Number after reversing the digits is 23. Sum = 32 + 23 = 55 = 11 x 5 Clearly the sum is divisible by 11. Again, consider the number 89. Number after reversing the digits is 98. Sum = 89 + 98 = 187 = 11 x 17 Clearly the sum is divisible by 11.
Q10. Write the condition to check the divisibility of a number by 5 and then identify the numbers from the following numbers which are divisible by 5.
2540, 4215, 423, 369, 78950, 7450, 457
Solution
A number is divisible by 5 if its unit's place digit is either 5 or 0.
From the given numbers, numbers divisible by 5 are:
2540, 4215, 78950, 7450.
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