When a number x is divided by 9, the remainder is 6. When the same number is divided by 21, the remainder is 12. If the x lies between 250 and 240, then what is the sum of all possible values of x.
Answers
Answer:
x needs to lie between 250-400(so that 2 numbers can be found to sum)
in that case sum would be 285+348=633
Step-by-step explanation:
We can form 2 equations one is 9k+6=x
other being 21 z+12=x
so
21z+12=x
=>(9*2+3)z+6+6=x
=>9*2z+6+(3z+6)=x
here 3z +6 must be divible by 9 so
3z+6=9
=>z=2
putting z=2
we get x=21*2+12=33
so 33 becomes the lowest value of x that gives remainder 6 when divided by 9 and 12 when divided by 21.
Now Lcm of 21 and 9 is 63
so the no. in between 250 and 400 must be number that is divisible by 63 to make it divisible by 21 and 9.
33 needs to be added to that number so that remainder when divided by 9 and 21 always remains 6 and 12 respectively.
so the no.s that are in between 250 and 400 are
252+33=285
315+33=348
their sum being 633
Hence the ans
Given : When a number x is divided by 9, the remainder is 6.
When the same number is divided by 21, the remainder is 12.
The x lies between 250 and 240
To Find : the sum of all possible values of x.
Solution:
x = 9A + 6
x = 21B + 12
9A + 6 = 21B + 12
=> 9A = 21B + 6
=> 3A = 7B + 2
B = 1 , A = 3 => x = 33
B =4 , A = 10 => x = 96
B = 7 , A = 17 => x = 159
and so on x = 222 , 285 , 348 , 411
There is no x between 250 and 240
If Question is between 250 and 400
Then x can be 285 , 348
Sum = 285 + 348 = 633
sum of all possible values of x is 633 if x is between 250 and 400
Learn More:
Find the number which when divided by 25 gives the quotient 20 ...
https://brainly.in/question/47028243
Find LCM and HCF of the following pairs of integers and verify that ...
brainly.in/question/17387230
Can 12 and 98 be HCF and LCM of two numbers - Brainly.in
brainly.in/question/17564109