•   Each digit is odd and number is divisible by 5 means the unit digit would be 5 (and not zero).

•   So, the number will be in the form ‘xy5’.

•   For the value of x and y we use remaining odd digits (1, 3, 7, 9) which can happen in 4 × 3 = 12 ways (as repetition of digits is now allowed).
• Thus there will be 12 such numbers.

This is another question from Data Sufficiency.

•   Question – Is x an integer

•   Statement 1 – x/3 is not an integer.

o  Suppose we take x = 4 (integer), then x/3 = 4/3 is not an integer.

o  But if we take x = 2/3 (not integer) then also x/3 = 2/9 is not an integer.

o  So, by using information provided in this statement we cannot say whether x is integer or not. Thus statement 1 alone is not sufficient.

•   Statement 2 – 3x is an integer.

o  If we take x = 4 (integer) then 3x = 12 is an integer. But if we take x = 1/3, then also 3x is an integer.

o  Thus, by using information provided in this statement alone we cannot say whether x is integer or not.

•   Combining both statement:

o  If we take x = 4 (integer) then x/3 is not integer and 3x = 12 is an integer. And if we take x = 1/3, then also x/3 is not an integer and 3x is an integer.

•   Thus, even combining both of the statements we can not say that whether x is an integer or not.

•   The product of two, two digits numbers has one 0 in it. That means the unit digit of one of the numbers must be equal to 5.

•   As the number is a two digit one then it will be of form 5_ × _ 5 = 2430.

•   To get zero the other digit must be an even number

•   Possible cases are 2 and 4

•   52 × 25 = 1300, 54 × 45 = 2430.

•   Thus digit is 54. Hence, 54 – 45 = 9

•   Statement is telling a possibility(use of ‘can’ in statement) thus while taking examples if even one case is like the statement then the statement will be true.

•   p is a prime number and q is a composite number. Lets assume p = 3, q = 9

•   Statement 1 – p × q = 3 × 9 = 27 is an odd number. So, statement 1 can be true.

•   Statement 2 - q/p = 9/3 = 3 is a prime number. So, statement 2 can be true.

•   Statement 3 – This statement is not true for same value of p and q as above however if we change the values for eq; p = 3, q = 4 then p+q = 7 which is a prime number, This statement becomes correct.

•   Thus answer is d.

•   Possibility I – (-)1, 0, 1, product = 0, sum = 0.

•   Possibility II – 1, 2, 3, product = 6, sum = 6.

•   Possibility III – -1, -2, -3, product = - 6, sum = - 6.

•   Only three such possibilities are there where sum=product.

•   Non zero digits= 9

•   Means X and Y can be replaced by 9 digits. So if X is replaced by 9 digits then Y is replaced by 8 digits (9-1)

•   This can be arranged at two places in 9 × 8 = 72 ways (without repetition).

•   three non-zero multiple of 3, digits are: 3, 6 and 9.

•   Note: a number more than 9 will make four-digit number. so, we are only taking these three.

•   the three-digit numbers can be--- 369,396,639,693,936,963

•   S= 369+396+693+639+936+963 = 3996

•   3996/9 = 444 (completely devisable).

•   3996/74= 54 (completely devisable).

•   So, both the statements are correct.

•   As, there are 5 consecutive integers, 5/100 =20.

•   Now we can check for consecutive integers preceeding and succeding 20 for sum 100.

•   ∴ 18 + 19 + 20 + 21 + 22 = 100 Thus, statement (1) is correct.

•   We know that, 1 + 2 + 3 = 6 And 1 × 2 × 3 = 6 So, Statement 2 is correct too.

•   Multiplying all options by 7, we get option (d): 15873 × 7 = 111111.

•  Here we do not need absolute values of p,q,r,s but an approximate idea to get the answer.

•  So, p – 2016 = q + 2017

or p = q + 2017 + 2016 (it means p > q)

•  q + 2017 = r – 2018

or r = q + 2017 + 2018 (it means r > q)

•  r – 2018 = s + 2019

or r = s + 2019 + 2018 (it means r > s)

•  p – 2016 = r – 2018

or r = p – 2016 + 2018 = p + 2 (it means r > p)

We already know that r > q, s

Hence, r is the largest number.

•  Since the digits are non repeating and we know that a prime number is a number which can only be completely divided by 1 and itself.

•  Adding the digits, we get a value of 15 which is divisible by 3.

•  Thus, any number formed by these 5 digits will always be divisible by 3.

•  Hence, there is no prime number without repeating the digits.

Let ® = x,

Then,

x + 1x + 5x + xx + x1 = 1xx  …(i)

Concept: write the numbers in the form of their place value.

Using (i), we have,

x + (10 + x) + (50 + x) + (10x + x) + (10x + 1) = 100 + 10x + x

=>24x + 61 = 100 + 11x

=>13x = 39

=> x = 3 Hence, ® = 3

•  Each letter represents a different digit greater than 3 => 4, 5, 6, 7, 8 & 9

  A  3  B  C

  D  E  2  F

1  5  9  0  2

•  Unit digit: C + F = 12

•  C, F may be 4 & 8 or 5 & 7 ----(1)

•  Ten digits: 1 (carry) + B + 2 = 0

•  So, B = 7 ----(2)

•  From (1) and (2) C & F will be 4 & 8

•  Hundred digits: 1 (carry) + 3 + E = 4 + E = 9

•  So, E = 5 ----(3)

•  Thousand digits: A + D = 15

•  A & D will be 6 & 9

•  Difference between A and D = 9 - 6 = 3

•  The integers between 1 and 100 which have 4 as a digit are:

4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84 and 94

•  So, there are a total 19 such integers.

•  Out of these, the integers which are divisible by 4 are:

•  4, 24, 40, 44, 48, 64 and 84

•  So, the number of integers not divisible by 4 = 19 – 7 = 12 integers

Multiplying the digit at ones place of each term in the product

1×7×5×2×5= 35×10= 350

Diving this by 100 we get remainder as 50

So the remainder when 51×27×35×62×75 is 50 when divided by 100.

Given number, N = (10n + 1)  Hence, sum of digits of number N will always be 2 if n = 0, 1, 2, 3.. or if n is any whole number.

From 1 to 1000, the numbers in which 5 can occur could be of one digit, two digits or three digits.

Case I – If the number is of one digit – 5 will appear only one time, i.e. in 5.

Case II – If the number is of two digits – then

(a) There is only one 5, this can happen in two ways _5 and 5_. In the first case (_5) the blank Place can be filled in 8 ways(as 0 and 5 cannot appear at that place), while in the second case (5_) the blank place can be filled in 9 ways (5 cannot appear there). Total 9 + 8 = 17 ways.

(b) There are two 5s. In this case only ONE possibility.

Case III – If the number is of three digits – then

(a) Only one 5. Then, 5 can occupy three positions. 5 _ _ or _ 5 _ or _ _ 5. In the first case (5_ _), remaining two positions can be filled in 9 way each. So total 9 × 9 = 81 possibilities. In the second case (_ 5 _) first position can be filled in 8 ways and last position can be filled in 9 ways. So total 9 × 8 = 72 possibilities. Same will be true for the third (_ _ 5) case. So total 72 possibilities.

(b) Only two 5. This can be done in three ways 55_ or 5_5 or _55. In first (55_) and second (5_5) case it can be filled in 9 ways each. While in the third case (_55) it can be filled in 8 ways. So total 9 + 9 + 8 = 26 possibilities.

(c) All three digits are 5. This can be done in only ONE way. i.e, 555.

So, total = 1 + 17 + 1 + 81 + 72 + 72 + 26 + 1 = 271. Ans.(b)

•  Let the ten's digit of two digit number ab be 'a' and unit’s digit be 'b'.

•  So the number will be of the form 10a + b.

•  After reversing the digits the number will be 10b + a.

•  By the condition given in question we have (10a+b)/(10b+a)=4/7 which means a/b=1/2

•  So by putting the values, total possible pairs (12, 21), (24, 42), (36, 63), (48, 84).

•  Thus, four pairs are possible.

•  Digits required to print one digit numbers (1 to 9) = 9×1 = 9

•  Digits required to print two digit numbers (10 to 99) = 90 × 2 = 180

•  Digits required to print three digit numbers (100 to 999) = 900 × 3 = 2700.

•  So, upto 999 pages we have 2700 + 180 + 9 = 2889 digits.

•  Now from here onwards each number will use 4 digits and we are remaining with 3089 - 2889 = 200 digits.

•  So 200/4 = 50 more numbers are there. i.e., 999 + 50 = 1049 pages in the book.

•  It is given 136 + 5B7 = 7A3.

•  Add the unit’s numbers to get 6 + 7 = 13. So, carry over 1.

=> 1 + 3 + B = 1A => 1 + 3 + B = 10 + A => B – A = 6.

•  Which means if A = 0, B = 6; if A = 1, B = 7, if A = 2, B = 8 and if A = 3, B = 9.

•  But given 7A3 is completely divisible by 3. So, as per rules of divisibility, 7 + A + 3 = 10 + A should also be completely divisible by 3.

•  So the possible values of A are 2, 5 and 8. (12 / 15 / 18 divisible by 3) Out of these, only 2 satisfies both the conditions so A = 2, so B = 8

•  We get 10 sets of triplets : (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1),(1,1,4), (1,4,1), (4,1,1), (2,2,2).

•  This is a question based on rules of divisibility.

•  The rule of divisibility of 3 is that the sum of all the digits of the number should be divisible by 3. We have 4 + 2 + 5 + 2 + 7 + 4 + 6 + B = 30 + B which is completely divisible by three.

•  So B can take value as 0, 3, 6, or 9. •  Hence 4 values are possible for B.

•   The given expression can be expanded as:

•   x + 10 + x + 20 + x + 10(x) + 3 + 10(x) + 1 = 210 + x

•   On solving, we get: x = 8.

•   Here we used the concept of face value and place value - a number 2• can be 25 = 20 + 5 = 2x10 + 5.

•   Let's assume  X=2 and  X=2 and =3

•   XY = 6, X/Y= 0.6, Y/X= 1.5, (X+Y)/XY= 5/6= 0.83

•   So, XY is the highest number.

•   Let the three-digits number be 100a + 10b + c.

•   b = 0

•   a + b + c = 4, or a + c = 4

•   100c + a = 100a + c + 198

•   99c - 99a = 198

•   99(c - a) = 198

•   c - a = 2. Ans is b

•   For numbers from 700 to 1000, the digit at hundred's place can be 7, 8 or 9. So, three possibilities are there.

•   Now, for the digit at ten's place to be greater than the digit at unit's place, total numbers is equal to the value of ten's digit. For example, for 2 at ten's digit, numbers can be 20, 21 (total 2). Similarly, for 6 at ten's place, numbers can be 60, 61, 62, 63, 64, 65 (total 6). 

•   When the first digit is 7, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 = 21

•   When the first digit is 8, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28

•   When the first digit is 9, total numbers are = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

•   Therefore, required total numbers = 21 + 28 + 36 = 85.

•   Statement 1 is wrong because values can be x= 5 and y= -3

•   Statement 2 is wrong because values can be x= 10, y= 2

•   Statement 3 is wrong because values can be x= -1, y= -9

•   These are consecutive odd numbers. The mean of the first 5 such numbers will be third such number, which is given to be as 39.

•   Hence, the series is: 35, 37, 39, 41, 43, 45, 47, 49, 51 and so on.

•   Now, mean of all the thirteen numbers = {[(13 -1)/2] + 1} th term = 7th term from the start, i.e. 47.

•   There are four such numbers: (168, 861) and (259, 952)

•   To find such numbers, first list down all 3 digits divisible by 7. This will generate a large set. In this set, remove those whose all digits are not different.

•   Among the remaining ones, look for the reverse digits pairings.

•   If R and S are multiple of 5, then R + S may or may not be divisible by 10.

•   For example, if R = 20 and S = 15. We will see that only (b) is correct.

From the given information, we have P → 4

Now, difference between P and T = 5

⸫  T— P = 5;  T — 4 = 5

]  T = 9

Now, again difference between N and T = 3

⸫  T — N = 3;  9 — N = 3  ] N = 6

Hence, option (d) is correct.