Elitmus previous questions - 6

Elitmus previous questions - 6

1. 99^n is such a number begin with 8, least value of n?

(a) 11
(b) 10
(c) 9
(d) n does not exist
Answer:
Explanation:
In a more traditional way, this problem can be solved like below.
99(100 - 1) = 9900-99= 9801
9801(100 - 1) = 980100-9801= 971299
971299(100 - 1) = 97129900 - 971299 = 96157601
.....
.....
Just observe the pattern, 98, 97, 96, .... for power of 2, 3, 4, .... So for 90 the power could be 10. So for 11, you get a number starts with 8.
Alternate method:
In a more elegant way, we can solve this question using logarithms.
For example,  log 90 = 1.9542, log 89 = 1. 9493.
Here characteristic is same as both numbers are two digit numbers.  Mantissa of 89 is less than mantissa of 90.
Similarly if you want to find a number starts with 8, it should be just less than a number starts with 9 and minimum.
⇒9.10x>99n$\Rightarrow {9.10}^x>{99}^n$
Suppose x = 1, the LHS = 90, for x = 2, LHS = 900.  So LHS is the least number starts with 9. and anything less than that number should starts with 8.
Let us take logarithm with base 10.
⇒log10(9.10x)>log10(99n)$\Rightarrow {\log }_{10}\left({9.10}^x\right)>{\log }_{10}\left({99}^n\right)$
⇒log109+log10(10x)>log10(99n)$\Rightarrow {\log }_{10}9+{\log }_{10}\left({10}^x\right)>{\log }_{10}\left({99}^n\right)$
⇒log109+x>n.log1099$\Rightarrow {\log }_{10}9+x>n.{\log }_{10}99$
Now the characteristic is not important.  We will take fraction part of the logarithm. { } represents fraction part of a number.
⇒log109>{n.log1099}$\Rightarrow {\log }_{10}9>\left\{n.{\log }_{10}99\right\}$
⇒$\Rightarrow$ 0.9542 > {n×1.9956}$\left\{n\times 1.9956\right\}$
For n = 11, we get 11×1.9956=21.9519$11\times 1.9956=21.9519$
So 0.9542 > 0.9519
So for n = 11, we get a number starts with 8.

2. Total 100 members are writing exam. In the 48 members are writing first exam. 45 members are writing second exam. 38 members are writing third exam. 5 members are writing all the three exams.  How many members are writing 2 exams?
Explanation:
Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5.  So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.

3. How many three digits no. can be formed also including the condition that the no. can have at least two same digits ?
Explanation:
Total number of 3 digit numbers = 9×10×10 = 900
Total number of numbers in which no digit repeats = 9×9×8 = 648
So the total number of numbers in which at least one digit repeats = 900 - 648 = 252

4. If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.
a) 36 
b) 24 
c) 18
d) 12
Answer: d
Explanation:
a,b,c are in G.P. so let the first term of G.P. = ar${\displaystyle \frac{a}{r}}$, and common ratio = r.
Therefore, a = ar${\displaystyle \frac{a}{r}}$, b = a$a$, c = ar$ar$
Given, loga+logb+logclog6=6${\displaystyle \frac{\log a+\log b+\log c}{\log 6}}=6$
⇒logabclog6=6$\Rightarrow {\displaystyle \frac{\log abc}{\log 6}}=6$
⇒log6abc=6⇒abc=66$\Rightarrow {\log }_{6}abc=6\Rightarrow abc=6^6$
put the value of a,b,c in gp format
⇒ar×a×ar=66$\Rightarrow {\displaystyle \frac{a}{r}}\times a\times ar=6^6$
⇒a3=66⇒a=36$\Rightarrow a^3=6^6\Rightarrow a=36$
Now a = 36r${\displaystyle \frac{36}{r}}$, b = 36, c = 36r.
We have to find the minimum value of c - b = 36r - 36.
r can be any number. So for r < 0, we get c - b negative.
When r = 1, c - b = 0
But none of the options are not representing it.
From the given options, r = 4/3, then c = 48. So option d satisfies this.

5. A natural number has exactly 10 divisors including 1 and itself.how many distint prime factors this natural number will have?
a. 1 or 2
b. 1 or 3
c. 1 or 2 or 3
d. 2 or 3
Answer: a
Explanation:
Number of factors of a number N$N$ is (p+1).(q+1).(r+1)... where N=ap×bq×cr...$N=a^p\times b^q\times c^r...$.
Given,  (p+1).(q+1).(r+1).. = 10.
From the above equation, p = 1, q = 4 or p = 9 satisfies.
So the number N is in the following two formats. a1×b4$a^1\times b^4$ or a9$a^9$
So it has either 1 or 2 prime factors.

6. How many values of c in x^2 - 5x + c, result in rational roots which are integers?
Explanation:
By the quadratic formula, the roots of x2−5x+c=0$x^2-5x+c=0$ are −(−5)±−52−4(1)(c)−−−−−−−−−−−√2(1)${\displaystyle \frac{-(-5)\pm \sqrt{-5^2-4(1)(c)}}{2(1)}}$ = 5±25−4c−−−−−−√2${\displaystyle \frac{5\pm \sqrt{25-4c}}{2}}$
To get rational roots, 25−4c$25-4c$ should be square of an odd number.  Why? because 5 + odd only divided by 2 perfectly.
Now let 25 - 4c = 1, then c = 6
If 25 - 4c = 9, then c = 4
If 25 - 4c = 25, then c = 0 and so on...
So infinite values are possible.