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Standard VI

Mathematics

Divisibility Rule for 3

Consider the ...

Question

Consider the number $N = 87 a 27931 b$ , where $b$ is a digit at unit's place and $a$ is a digit at ten lakh's place. Answer the following questions.
Number of values of $a + b$ for which $N$ is divisible by $11$ is

0

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Solution

The correct option is D $2$
$N = 87 a 27931 b$

$t o m a k e N d i v i s i b l e b y 11$

$b - 1 + 3 - 9 + 7 - 2 + a - 7 + 8 i s d i v i s i b l e b y 11$

$= a + b - 19 + 18$ $= a + b - 1$

$i f a + b = 1 t h e n s u m = 0, a n d t h u s$

$N i s d i v i s i b l e b y 11$

$∴ a + b = 1$

$I f a + b = 12 t h e n s u m = 11, a n d t h u s N i s d i v i s i b l e b y 11$

$S o, a + b = 2 o r 12$

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Similar questions

Q. Consider the number

N = 87 a 27931 b

, where

b

is a digit at unit's place and

a

is a digit at ten lakh's place. Answer the following questions.
The least value of

a

for which

N

is divisible by

12

Question 13

In the following question out of the four options only one is correct, write the correct answer.

The digit in the ten's place of a two-digit number is 3 more than the digit in the unit's place. Let the digit at unit's place be b. Then, the number is:

(a)11b + 30

(b) 10b + 30

(d) 10b + 3

Question 167

In a five-digit number, digit at ten's place is 4, digit at unit's place is one - fourth of ten's place digit, digit at hundred's place is 0, digit at thousand's place is 5 times of the digit at unit's place and ten thousand's place digit is double the digit at ten's place. Write the number.

Q. In a five-digits number, digit at ten's place is

3,

digit at unit's place is one third of ten's place digit, digit at hundred's place is

0,

digit at thousand's place is

4

times of the digit at unit's place and ten thousand's place digit is triple the digit at ten's place. Then the number is:

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Consider the number N=8 7 a 2 7 9 3 1 b, where b is a digit at unit's place and a is a digit at ten lakh's place. Answer the following questions. Number of values of a+b for which N is divisible by 11 is

The correct option is D 2N=87a27931btomakeNdivisibleby11b−1+3−9+7−2+a−7+8isdivisibleby11=a+b−19+18=a+b−1ifa+b=1thensum=0,andthusNisdivisibleby11∴a+b=1If a+b=12 thensum=11, and thus N is divisible by 11So,a+b=2 or 12

Consider the number $N = 87 a 27931 b$ , where $b$ is a digit at unit's place and $a$ is a digit at ten lakh's place. Answer the following questions.
Number of values of $a + b$ for which $N$ is divisible by $11$ is