Numbers Questions

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

---> 2 x 3 x 5 = 30 

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.

Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.

We cannot have three digits as (3, 3, 2) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

A perfect cube will have prime factors that are in groups of 3; for example 125 has the prime factors 5 x 5 x 5 , and 64 x 125 will also be a cube because its factors will be 4 x 4 x 4 x 5 x 5 x 5

Consider the answer choices in turn.

8 is the cube of 2, and p is a cube, and so the product will also be a cube.

pq will also be a cube as shown above.

pq is a cube and so is 27, but their sum need not be a cube. Consider the case where p =1 and q = 8, the sum of pq and 27 will be 35 which has factors 5 x 7 and is not a cube.

-p will be a cube.

Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).

When 16nis divided by 9, we have

 1619, remainder = 7  

1629, remainder = 4 

1639, remainder = 1 

1649, remainder = 7 

1659, remainder = 4 

1669, remainder = 1 

So, we have cyclicity of 3 factors i.e 7,4,1. 

Hence only 3 remainders are possible.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.

In case B, odd times even will give even

1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320