Quiz 5

Both the sequences ( 17, 21, 25 ……………… and (16, 21, 26….. are arithmetic progression with a common difference of 4 and 5 respectively.
In both the sequence first common term is 21.
Hence a new arithmetic sequence containing the common terms of both the series can be formed with a common difference of LCM of ( 4, 5) is 20
New sequence will be 21, 41, 61, ….401
n^th term = a + (n-1)d
401= 21 + (n-1)20
n-1=19
Hence, n=20
(b)

The number required is greater than 999 and less than and equal to 4000.
Now out off our digits, 0, 1, 2, 3, 4.
To form a number greater than 999 and less than 4000.
The digit at thousands place can be selected in 3 ways( 0 and 4 cannot be taken )
The digit at hundreds place can be selected in 5 ways.
The digit at tens place can be selected in 5 ways.
Total required number of ways= 3 x 5 x 5 x 5 = 375 ways
Since, 4000 is also one of the required number.
Therefore, total number of ways=375 + 1
(d)

Number of terms in the given expansion is nothing but the non-negative integral solutions of the equation:

a + b + c = 20

Total number of non-negative integral solutions

^(20+3-1)C_(3-1) = ²²C₂ = 231

According to question, if two numbers say a and b are erased and replaced by a new number a + b – 1, then in every repetition, the number of integers gets reduced by 1 and consequently at the last repetition there will be only one number left.
Whatever may be our selection of two numbers a and b. In any and every repetition, the final number so arrived will not changes.
Now, the sum of integers from 1 to 40= n(n + 1)/2 = 820
As, discussed above the sum of integers of the first, second, third ……… repetitions will be 819, 818, 817, ………….. so on respectively. Therefore, after 39 operations there will be only 1 number left and that will be 820 – 39=781

Answer - c

A total number of two-digit codes that can be formed is 10 x 10 = 100.
Out of the 0, 1, 6, 8, 9 can create confusion.
Using these five digits total number of two-digit numbers that can be made is 5 x 5 = 25.
But out of these 25 numbers 00,11,88,69 and 96 will not make any confusion.
Hence answer is 100 - 25 + 5 =80

Thumb Finger :1, 9, 17, ….
Index Finger: 2, 8, 10, 16, 18, ….
Middle Finger: 3, 7, 11, 16, 19, ….
Ring Finger : 4, 6, 12, 14, 20, ……
Little Finger : 5, 13, 21, ….
Numbers on thumb forms a AP with common difference=8
Numbers on middle forms a AP with common difference=4
1993 will be on thumb.
Hence, 1994 will be on index finger.

U(n+1)= 2Un + 1, ( n=0, 1, 2, …………) U0=0
Put
n=0, U1=1
n=1, U2=3
n=2, U3=7
n=4, U5=31
Seeing this pattern we can conclude i.e. U(n)= 2ⁿ -1
Hence U(10)= (2)¹⁰ – 1= 1023

The product of all integers from 1 to 100 is 100!.

The number of trailing zeros in 100! is (100/5)+(100/25) = 24

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

There are 50 odd numbers less than 100 which are not divisible by 2. Out of these 50 there are 17 number which are divisible by 3.
Out of remaining there are 7 numbers which are divisible by 5.
Hence, numbers which are not divisible by 2, 3, 5 = 50 – 17 – 7 = 26