Lakshya Education MCQs

Question: N= 4! + 5!+6!.........200!. Find the last two digits of N
Options:
A.00
B.13
C.04
D.54
E.24
Answer: Option C
: C

After 10!, the factorials end with 0 as the last two digits. We thusneed to find the last two digits from 4! To 9! Only 4!+5!+6!....9!= 24+120 + 720 + 5040 + 40320 + 362880 = ___04

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Question 1. How many two digit odd numbers are there with 10 factors?
  1.    1
  2.    2
  3.    3
  4.    4
  5.    None of these
Answer: Option E
: E

10 factors implies that the number can be of the form 1) a9 ( Least odd number will be 39) 2) ab4 (Least odd number will be 345=405) Therefore, as the least number is 405, there are no two digit numbers of the specified form. The answer is 0. If a number N can be factorized as N=ambn(a, b are the prime factors) Then number of factors= (m+1)(n+1)
Question 2. What is the value of m , if a3+3a2am+4 when divided by a2, gives a remainder m?
  1.    20
  2.    8
  3.    0
  4.    4
  5.    1
Answer: Option B
: B

We know that Number = Quotient × Divisor + Remainder If Divisor (a-2) equals 0, then Number = Remainder. This is the property we are going to use Make the divisor=0; i.e. put a=2, Then a3+3a2am+4=m where a=2 Thus, 3m= 24 and m=8.
Question 3. Find the least number which when divided by 9, 10 and 11 give remainders of 2,3 and 5 respectively.
  1.    1523
  2.    3543
  3.    992
  4.    543
  5.    none of these
Answer: Option E
: E

Use Chinese remainder theorem, Let the number be N. N is of the form 9A+2 (i.e. N divided by 9 gives a remainder 2)= 10B+3 (i.e. N divided by 10 gives a remainder 3) =11C+5 (i.e. N divided by 11 gives a remainder of 5) We will first find the number which is of the form 9A+2=10B+3.--------(1) Here for B=8, A=9, which are the first integer solutions. Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3. The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2..... Now to include the 3rd condition(i.e divisor 11 and remainder 5) we can write 83+90k=11C+5 -------(2) Find the first integer value of K which satisfies the equation such that C is also an integer. Here for K = 5, C = 48, which are the first integer solutions. Substituting this in equation (2) we get the value 90 (5) + 83= 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5. Shortcut:- You may be tempted to mark option (a) after a first glimpse at the options. Since the question asks for the least number, you should be careful while answering. You can still back calculate using answer option (a) option (a) 1523 satisfies the condition. Hence 1523-990 = 533 has to be the first such number. Answer is option (e) Also note that, the first number has to be < 990.
Question 4. Find the 202nd digit from the right in the product of 4!×5!×6!...............71!?
  1.    1
  2.    2
  3.    3
  4.    0
  5.    4
Answer: Option D
: D

Note that 71! has 16 zeroes at its end.(You can find this out by finding the highest power of 5 in 71!) So does 70!. 66! To 69! Have 15 zeroes at its end. Multiplication of these itself amounts to >202 Hence the 202nd digit is 0 To find the highest power of a number in a factorial a)Highest power of a prime number in a factorial: To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients. Eg) Find the highest power of 100! Solution: 1005=20;205=4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24 b)Highest number of a composite number in factorial 1)Factorize the number into primes. 2)Find the highest power of all the prime numbers in that factorial using the previous method. 3)Take the least power.
Question 5. Find the number of 4 digit numbers which are divisible by 11 or 13 but not by 17?
  1.    1133
  2.    1363
  3.    1455
  4.    1463
  5.    1123
Answer: Option B
: B

Following are the steps while calculating the answer 4 digit numbers divisible by 11=899911=819,4 digit numbers divisible by 13=899913=692 or 693 Number of integers common to 11 and 13 (as questionis 11 or 13) =8999(13×11)=62 or 63 To subtract numbers which are common to 11&17;13&17;11,13&17 11&17=8999(11×17)=47 or 48 terms 13&17=8999(13×17)=40 or 41 11,13,17=8999(11×13×17)=3 or 4 terms Total ( you can approximate as the answers are far apart)= 819+693-63-48-41+4= 1364 Closest answer is option (2)
Question 6. If (127)8+(12)6=(x)5. Find the value of x.
  1.    230
  2.    124
  3.    340
  4.    240
  5.    None of these
Answer: Option C
: C

Write all the numbers in base 10 and then arrive at the answer 127 in base 8=7×1+2×8+1×82=87 in base 10 12 in base 6=2×1+1×6=8 in base 10 Adding both =95 in base 10 Expressing in base 5=340

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