Question

If ‘ abca ’ is a 4-digit number which equals to
a^b x c^a .
Find a, b and c

Answer 1

1000a+100b+10c+a = a^b * c^a
1001a+100b+10c = a^b * c^a

Answer 2

i think it should be abcd not abca

Answer 3

wait i got it

Answer 4

LHS is divisible by 'a', so RHS must also be divisible by 'a'.

Therefore, let (100b + 10c) / a = J (where J is an integer), so,
[2] 10(10b + c) = aJ

The factors of this LHS include 1, 2, 5, 10, (10b + c) and
any factors of (10b + c).

If 'a' is a single digit, it can only be 2, 5, or a factor of
(10b + c). (note that 'a' cannot be 1)
I'm going to dismiss the factors of (10b + c) as I haven't
a clue what they are.

First, let a = 5.
Therefore, from [2] we get : 100b + 10c = 5J
Plugging these into [1] gives :
(5^b)(c^5) = 5005 + 5J

Dividing through by 5 :
[3] [5^(b-1)](c^5) = 1001 + J

Now, 100b + 10c = 5J is a minimum of 110 if b = 1, c = 1,
and is a maximum of 990 if b = 9, c = 9.
Thus, 110 <= 5J <= 990, or, 22 <= J <= 198.

So in [3], 1023 <= [5^(b-1)](c^5) <= 1199
Testing shows that c cannot be 1, 2, 3 or >4,
but it can be true for c = 4, giving b = 1,
but, [1] is then not true. Therefore, 'a' is not equal to 5.

So try a = 2.
In [1], we have : (2^b)(c^2) = 2002 + 100b + 10c
Dividing by 2 :
[2^(b-1)](c^2) = 1001 + 50b + 5c
Minimum value for 1001 + 50b + 5c is 1056.
Maximum value is 1496.
So, 1056 <= [2^(b-1)](c^2) <= 1496
Because of c^2, we must be able to take the square root.
Therefore, 32.5 <= {2^[(b-1)/2]}c <= 38.7
Trying values for c, it's found that only 2^2 * 9 = 36, works.
Thus, c = 9 and (b-1)/2 = 2, so b = 5.

Answer 5

@sauravshakya

Answer 6

BUT 2592 is not equal to 2^5 * 9^2

Answer 7

i m not sure

Answer 8

2592 is equal to (2^5)*(9^2).\[2^5\cdot 9^2=32\cdot 81 = 2592\]

Answer 9

oh yes....
thought 9^2 =18

Answer 10

my answer is correct..i was in doubt it is wrong solved by me.
thank u so much

Answer 11

Thank u so much @sauravshakya and @joemath314159