Number System
- The total number of integers between 100 and 200, which are divisible by both 9 and 6, is :
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A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18.
Correct Option: B
A number is divisible by 9 and 6 both, if it is divisible by LCM of 9 and 6 i.e., 18. Hence, the numbers are 108, 126, 144, 162, 180, 198.
Thus , the total number of integers between 100 and 200 are 6.
- The least number, which must be added to 6709 to make it exactly divisible by 9, is
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A number is divisible by 9 if the sum of its digits is divisible by 9.
Here, 6 + 7 + 0 + 9 = 22Correct Option: A
A number is divisible by 9 if the sum of its digits is divisible by 9.
Here, 6 + 7 + 0 + 9 = 22
Now, 22 + 5 = 27, which is divisible by 9. Hence 5 must be added to 6709.
- How many 3-digit numbers, in all, are divisible by 6 ?
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First 3–digit number divisible by 6 = 102
Last such 3-digit number =996
∴ 996 = 102 + (n –1) 6
⇒ (n – 1)6 = 996 – 102 = 894Correct Option: B
First 3–digit number divisible by 6 = 102
Last such 3-digit number =996
∴ 996 = 102 + (n –1) 6
⇒ (n – 1)6 = 996 – 102 = 894⇒ n – 1 = 894 = 149 6
⇒ n = 150
Hence , required answer is 150.
- (49)15 – 1 is exactly divisible by :
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As we know that ,
xn – an is exactly divisible by (x – a) if n is odd.Correct Option: D
As we know that ,
xn – an is exactly divisible by (x – a) if n is odd.
∴ (49)15 – (1 )15 is exactly divisible by 49 – 1 = 48, that is a multiple of 8.
Hence required answer is 8.
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1 + 999 + 692 × 99 is equal to 7 693
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Expression = 1 + 999 + 692 × 99 7 693
Correct Option: B
Expression = 1 + 999 + 692 × 99 7 693 = 1 + 999 × 99 + 692 × 99 7 693 = 1 + (1000 − 1) 99 + 692 7 7 = 1 + 692 + 99000 − 99 7 7 = 693 + 99000 − 99 7
Required answer = 99 + 99000 – 99 = 99000