Reverse the three digit number and simplify it:-
A) 872
B) 232
C) 401
D) 512
E) 723
F) 732
G) 514
H) 718
I) 234
Answers
Answer:
Let the number that’s divisible by 7 (normally and after reversing) be abc, which is nothing but 100a + 10b + c.
Since 98a + 7b is divisible by 7, we can subtract it from the original number, without any loss of generality, so as to keep the original number intact from the divisibility by 7 point of view.
Thus, 100a+10b+c - (98a+7b)
= 2a + 3b + c
And since we assumed that abc was divisible by 7, we can assert that 2a + 3b + c is also divisible by 7.
Since we are looking for a number that’s also divisible by 7 after reversing, we have 100c + 10b + a is also divisible by 7 and thus 100c + 10b + a - (98c + 7b) is also divisible by 7. This is nothing but 2c + 3b + a and this is also divisible by 7.
So, we have 2 equations that are divisible by 7, as per below:
2a + 3b + c = 7x ————————- I
2c + 3b + a = 7y ————————- II
Multiplying Equation II by 2 and and subtracting I from it, we get:
4c + 6b + 2a - 2a - 3b - c
= 3b + 3c
Since we have subtracted a number that’s divisible by 7 from another number that’s divisible by 7, the outcome is also divisible by 7 and thus 3b + 3c is divisible by 7.
Now, this is possible only if 3*(b + c) is divisible by 7, which translates into (b+c) being divisible by 7, since 3 and 7 are co-prime numbers (they are prime numbers anyway).
Here, we are in a situation where we have more variables (a, b and c) and fewer equations (just 2 equations, I and II) and this means there are possibly more solutions and also that we will have to proceed in “Trial and Error” fashion.
Since we know that (b+c) is divisible by 7, we have either b+c = 7 or b+c = 14.
Proceeding with Trial and Error approach, we will have to substitute with b = 0 .. 9 and accordingly set appropriate values for c and then try out either equation I or equation II and find possible value or values of a.
We get many solutions but in all but 2 solutions, we are getting a = c and this is not acceptable as per given condition.
The 2 solutions that we get are:
When b = 5 and c = 2, we get a = 9and
When b = 6 and c = 1, we get a = 8.
Thus the required pairs are 952, 259 and 861, 168.
Both 952 and 259 are divisible by 7 (136*7 = 952 and 37*7 = 259) and similarly both 861 and 168 re divisible by 7 (123*7 = 861 and 24*7 = 168
Answer:
A) 278
B) 232
C) 104
D) 215
E) 327
F) 237
G) 415
H) 817
I) 432
Step-by-step explanation:
Reversing the number ⤵
872➡278
2 is given last so we write it first the 7 is followed by it then we write 7 after that we write 8 because is given first .
Similarly all.
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Hope this is helpful to you!