# Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 2)

2.

(112 x 5

^{4}) = ?Answer: Option

Explanation:

(112 x 5^{4}) = 112 x |
10 | 4 | = | 112 x 10^{4} |
= | 1120000 | = 70000 | ||

2 | 2^{4} |
16 |

Discussion:

121 comments Page 1 of 13.
BedaFabio said:
8 years ago

112*5^4=112*(10/2)^4

= 2*56*(10^4)/(2^4)

2 * 2 * 28 * 10^4

-------------------

2^4

2 * 2 * 2 * 14 * (10 * 10 * 10 * 10)

--------------------------------

2 * 2 * 2 * 2

2 * 2 * 2* 2 * 7 * 10 * 10 * 10 * 10

-----------------------------------

2 * 2 * 2 * 2

Then simply remove the number which is above and below the " ------------" bar

to simply the fraction till any above's number will not repeat( present ) below

thus, we get:

112 * 5^4 = 7 * 10 * 10 * 10 * 10 = 7 0 0 0 0.

That's all!

= 2*56*(10^4)/(2^4)

2 * 2 * 28 * 10^4

-------------------

2^4

2 * 2 * 2 * 14 * (10 * 10 * 10 * 10)

--------------------------------

2 * 2 * 2 * 2

2 * 2 * 2* 2 * 7 * 10 * 10 * 10 * 10

-----------------------------------

2 * 2 * 2 * 2

Then simply remove the number which is above and below the " ------------" bar

to simply the fraction till any above's number will not repeat( present ) below

thus, we get:

112 * 5^4 = 7 * 10 * 10 * 10 * 10 = 7 0 0 0 0.

That's all!

Phani Gorantla said:
1 decade ago

Here we can apply 2*5 as a zero technique

for ex 2*2^2*5^2

Now ans is 200(i.e in this pbm two 2's two 5 combination is there thats why i just put two zeros after 2.

Now the given pbm i converted into 2's and 5's combination

(112)*5^4

(7*2^4)*5^4

i.e 7*(2^4*5^4)ans is 70000(4 zeros added after 7 because four 2's and 5's combinations are there)

for ex 2*2^2*5^2

Now ans is 200(i.e in this pbm two 2's two 5 combination is there thats why i just put two zeros after 2.

Now the given pbm i converted into 2's and 5's combination

(112)*5^4

(7*2^4)*5^4

i.e 7*(2^4*5^4)ans is 70000(4 zeros added after 7 because four 2's and 5's combinations are there)

Umar kce said:
9 years ago

Our author says easiest way.

Ex: Any no which is multiply by multiples of 5 we can follow these method.

= 187*5= 1870/2 = 935.

i.e, 5^1 = 5. So one zero.

If 125 means 5^3 = 125, so 000 to be added.

How many zero to be added in numerator that much of 2 to be added in denominator?

If 625 means in 1870000/2*2*2*2.

Ex: Any no which is multiply by multiples of 5 we can follow these method.

= 187*5= 1870/2 = 935.

i.e, 5^1 = 5. So one zero.

If 125 means 5^3 = 125, so 000 to be added.

How many zero to be added in numerator that much of 2 to be added in denominator?

If 625 means in 1870000/2*2*2*2.

Deepeshsingh said:
1 decade ago

I did not multiply these numbers to get the answer. The simplest method is to find the number of zeros. See here we have four 5 and we can have fours 2, which when multiplied will give maximum number of fours zeros. And there is only one option with 4 0's that is 7000. :).

Ravi said:
1 decade ago

Hi @Ritesh, Its simple man: your Q: 10/2.

You can divide this number 10/2 you will get answer 5.

We just take that form (10/2) for our convenient. Moreover you can take also 50/2 if you take, would be 25. That's all buddy.

You can divide this number 10/2 you will get answer 5.

We just take that form (10/2) for our convenient. Moreover you can take also 50/2 if you take, would be 25. That's all buddy.

RAVI KANT said:
9 years ago

Multiplication of a number by 5^n: Put n zeros to the right of the multiplicand and divide the number so formed by 2^n.

So (112 x 5^4) = 1120000/2^4 = 1120000/16 = 70000.

Note: It is only apply for 5^n.

Thank you.

So (112 x 5^4) = 1120000/2^4 = 1120000/16 = 70000.

Note: It is only apply for 5^n.

Thank you.

(1)

Aniruddha Wagh said:
9 years ago

2x5 gives 0 at unit place.

2^2x5^2 gives two zeros at last of number i.e. at unit and tens place.

Similarly for higher power gives respective zeroes at end. Now you can easily solved it. No need for me to solve.

2^2x5^2 gives two zeros at last of number i.e. at unit and tens place.

Similarly for higher power gives respective zeroes at end. Now you can easily solved it. No need for me to solve.

G.manoj said:
5 years ago

5^4 = 5 * 5 * 5 * 5, 5 * 5 = 25.

So, 25 * 25 = 625 we can split 112 = 100 + 10 + 2 then;

100 * 625 = 62500,

10 * 625 = 6250,

2 * 625 = 1250 and finally answer is 62500 + 6250 + 1250 = 70000.

So, 25 * 25 = 625 we can split 112 = 100 + 10 + 2 then;

100 * 625 = 62500,

10 * 625 = 6250,

2 * 625 = 1250 and finally answer is 62500 + 6250 + 1250 = 70000.

(1)

Ramu Nakka said:
7 years ago

In simple Method we can solve this problem Like:

5 to the power of 4 means 5*5*5*5 = 625.

You can (112*5^4),

= 112*(5*5*5*5),

= 112*(625),

= 70000.

5 to the power of 4 means 5*5*5*5 = 625.

You can (112*5^4),

= 112*(5*5*5*5),

= 112*(625),

= 70000.

Pratheev Kalyan said:
7 years ago

(112)*(5^4)=(2*2*2*2*7)*(5*5*5*5).

=(2*5)*(2*5)*(2*5)*(2*5)*7,

=10*10*10*10*7,

=1000*7,

=70000(ans).

=(2*5)*(2*5)*(2*5)*(2*5)*7,

=10*10*10*10*7,

=1000*7,

=70000(ans).

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