Numbers Questions

Let the first number be K and second number be L
Then, 2K + 3L = 100.....(1)
3K + 2L = 120.......(2)
solving (1)&(2), we get
K = 32 and L = 12

Hence the largest number is 32.

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).

Let y be the total number of loaves
a be the remaining loaves after 1st thief
b be the remaining loaves after 2nd thief
and c, d..so on after 3 and 4 th
e = 0 be the remaining loaves after 5th thief

a = y - (y/2 + 1/2)
= y/2 - 1/2

b = a - (a/2 + 1/2)
= a/2 - 1/2

c = b - (b/2 + 1/2)
= b/2 - 1/2

d = c - (c/2 + 1/2)
= c/2 - 1/2

e = d - (d/2 + 1/2)
= d/2 - 1/2
we know that, e = 0,
thus by substitution, we get
d = 1
c = 3
b = 7
a = 15
y = 31
Therefore total number of bread loaves in bakery was 31.

Smallest number of 6 digits is 100000

on dividing 100000 by 111 we get 100 as remainder 

 Number to be added = (111 -100) = 11

 Required Number  = 100011

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.

Divisor = 

           = 

To solve a pair of simultaneous equations such as those given we can add or subtract them.

Adding we get 4x + 4y = 20

Therefore 2x + 2y = 10