[krsh.vik]

Originally Posted by Destiny's_Child

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Quant Question # 12
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Let 123abc231bca312cab be a 18 digit number in base 7. How many ordered pairs (a,b,c)
exist such that the given 18 digit number is divisible by 4?
(a) 62 (b) 74 (c) 86 (d) none of these

(49c + 7a + b)mod 100 should be divisible by 4
Is there any easier method than finding all the possibilities ??

[Aarav]

Originally Posted by krsh.vik

(49c + 7a + b)mod 100 should be divisible by 4
Is there any easier method than finding all the possibilities ??

In base 10, we know the divisibility criteria by 11. Please reason this out, as to why that should be like that only. In any base b the divisor of (b+1) divides the number iff the sum of the digits with alternating sign divides that divisor of (b+1). From here, the calculations get relatively easier.

E.g. 1122 or 1221 is divisible by 13 and 26 in base 25. And 1122 or 1221 is divisible by 3, 9 and 27 in base 26.

Here, 1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b => 6 -(a +b +c) is divisible by 4

[Aarav]

Solution to the problems. Will request you all to read #10 and #12 well.

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Quant Answer # 10
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Let us denote S(n) as the sum n of a, b, c, d the 4 elements from the
set such that 1 <= a < b < c < d <= 100.

Consider the statement (1)
S(n) < 201 => a+b+c+d < 201 => (101-p) + (101-q) + (101-r) + (101-s) < 201
=> P+q+r+s > 203.

**Note: We are subtracting 101 from each element because for each "a"
belonging to set A, "101-a" also belongs to the set A.

Thus, for each set {a,b,c,d} where S(n) < 201 we have an exact map
where S(n) > 203.
=> the number of 4 element subset that have sum < 201 is same as number of 4 element subset that have sum > 203.

But for S(n)=202, we have few extra sets of {a,b,c,d}.
Hence, (1) is true.


Consider the statement (2)
S(n) > 203 => a+b+c+d > 203 => (101-p) + (101-q) + (101-r) + (101-s) > 203
=> p+q+r+s < 201

Thus, for each set {a,b,c,d} where S(n) > 203 we have an exact map
where S(n) < 201.

=> the number of 4 element subset that have sum > 203 is same as number of 4 element subset that have sum < 201.

But for S(n)=202, we have few extra sets of {a,b,c,d}.
Hence, (2) is false.

Hence, choice (a) is the correct answer.


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Quant Answer # 11
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2f(x) + f(1-x) = 2*x^2 + 1, (1)
2f(1-x) + f(1-(1-x)) = 2f(1-x) + f(x) = 2*(1-x)^2 + 1, (2)
subtracting 2nd from 1st we get, f(x) - f(x-1) = 4x -2, (3)

adding 3rd with the first we get, 3f(x) = 2*x^2 + 4x - 1

thus f(x) = 1/3(2*x^2 + 4x - 1) = 1/3(2(x+1)^2 - 3)
which assumes the min. value at x = - 1 and f(x) >= -1.

Hence, choice (b) is the correct answer.


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Quant Answer # 12
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Any number in base k is divisible by k+1 or its divisors if and only if the sum of the digits(alternating and sign changed) is divisible by k+1 or its divisors respectively.

eg. in base 10, if a number is div. by 11 then sum of its digits(sign changed alternatively) is div. by 11 and vice-versa.

In base 7 k = 7, k+1 = 8. 4 is the divisor of 8.
The alternating sum of the digits of 123abc231bca312cab is
1-2+3-a+b-c+2-3+1-b+c-a+3-1+2-c+a-b => 6 -(a +b +c) is divisible by 4


what values can a+b+c take?
a+b+c = 2, (0,1,1), (0,0,2), 3+3 = 6 solutions
a+b+c = 6, (0,0,6), (0,1,5), (0,2,4), (0,3,3), (1,1,4), (1,2,3), (2,2,2), 3+6+6+3+3+6+1 = 28 solutions
a+b+c = 10, (1,3,6), (1,4,5), (2,2,6), (2,3,5), (2,4,4), (3,3,4), (4,0,6), (5,0,5), 6+6+3+6+3+3+6+3 = 36 solutions
a+b+c = 14, (2,6,6), (3,5,6), (4,4,6), (4,5,5), 3+6+3+3 = 15 solutions
a+b+c = 18, (6,6,6), 1 solution

In all we have 86 ordered pairs of (a,b,c)

Hence, choice (c) is the correct option.

[Aarav]

Ok ... here are 3 more. By this time I have started believing that these problems have been already solved by majority. Can't be that junta is lazy?

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Quant Question # 13
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If |x^2-1|/(x-2) = x, then what value can x take?

(a) (1-5^1/2)/2 (b) (1-2^1/2)/2 (c) (1-3^1/2)/2 (d) (1+2^1/2)/2


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Quant Question # 14
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Two brothers, each aged between 10 and 90, "combined" their ages by writing them down one after the other to create a four digit number, and discovered this number to be the square of an integer. Nine years later they repeated this process (combining their ages in the same order) and found that the combination was again a square of another integer. What was the sum of their original ages?

(a) 37 (b) 55 (c) 63 (d) 71


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Quant Question # 15
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If f(n) = (24n + 1)^1/2, where n is a positive integer,
then what can be said about f(n) ?

(a) f(n) is prime for some n, but not infinite n
(b) f(n) is composite for some n, but not infinite n
(c) both (a), (b) are true
(d) both (a), (b) are false

[krsh.vik]

Originally Posted by Aarav

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Quant Question # 15
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If f(n) = (24n + 1)^1/2, where n is a positive integer,
then what can be said about f(n) ?

(a) f(n) is prime for some n, but not infinite n
(b) f(n) is composite for some n, but not infinite n
(c) both (a), (b) are true
(d) both (a), (b) are false

24n+1 would definitely yield all squares of prime numbers
Answer D

[krsh.vik]

Originally Posted by Aarav

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Quant Question # 13
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If |x^2-1|/(x-2) = x, then what value can x take?

(a) (1-5^1/2)/2 (b) (1-2^1/2)/2 (c) (1-3^1/2)/2 (d) (1+2^1/2)/2

suppose x>=1
x^2-1=x^2-2x
x=1/2 not correct
suppose x<1
1-x^2=x^2-2x
2x^2-2x-1=0
roots
2(+/-)root(12)/4
2(+/-)2root3/4
1(+/-)root3/2
ans c

[krsh.vik]

Originally Posted by Aarav

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Quant Question # 14
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Two brothers, each aged between 10 and 90, "combined" their ages by writing them down one after the other to create a four digit number, and discovered this number to be the square of an integer. Nine years later they repeated this process (combining their ages in the same order) and found that the combination was again a square of another integer. What was the sum of their original ages?

(a) 37 (b) 55 (c) 63 (d) 71

suppose sum of the ages is 37
Then add them 37 times
after 9 years add them 55 times option A satisfies
similarly for all the options ...
Am I missing something here ???

[Aarav]

Originally Posted by krsh.vik

suppose sum of the ages is 37
Then add them 37 times
after 9 years add them 55 times option A satisfies
similarly for all the options ...
Am I missing something here ???

If the ages were 37 and 21, them combining means 3721 and it's a perfect square. After 9 years the number will be 4630 - it's not a perfect square here, but question wants it like that.

[krsh.vik]

Originally Posted by Aarav

If the ages were 37 and 21, them combining means 3721 and it's a perfect square. After 9 years the number will be 4630 - it's not a perfect square here, but question wants it like that.

answer is a
2116 after nine years 3025

[Aarav]

The attendance on weekends is generally low compared to weekdays. Vineet will edit the rules of the thread. We will discuss the questions here only on weekdays. I think that should be fine with everyone?

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Quant Answer # 13
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|x^2-1|/(x-2) = x => |x^2-1| = x(x-2). Since LHS >= 0 RHS = x(x-2) >= 0
=> x >= 2, x <= 0.

case when x >= 2:
|x^2-1| = x(x-2) => x^2 - 1 = x^2 - 2x => x=1/2, rejected

case when x <= 0:
In this we should have 2 sub-cases as well since |x^2-1| = x^2-1 for x <= -1 and 1-x^2 for -1 <= x <= 0.

when x <= -1 we have x^2 - 1 = x^2 - 2x => x=1/2, rejected
when -1 <= x <= 0, we have 1- x^2 = x^2 -2x => 2x^2 -2x -1 = 0

solving we get (1-3^1/2)/2 and (1+3^1/2)/2. The later value is rejected as it's > 0.

Hence, choice (c) is the right answer.



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Quant Answer # 14
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Let x, y be the ages. Then 100x + y = a^2, also
100(x + 9) + b + 9 = b^2, subtracting we get
b^2 - a^2 = 909 = 101*9 = 303*3 = 909*1
thus, (b,a) are (55, 46) (156, 150), (455, 454)
rejecting the last 2 pairs as a <= 99, we get (b, a) as (55, 46)

a^2 = 2116, thus the sum of the ages is 21 + 16 = 37

Hence, choice (a) is the right answer.



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Quant Answer # 15
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We know that every prime > 3 can be expressed as 6k+/-1, where k is a natural number.
Let p = 6k+/-1, then p^2 = 36k^2 +/-12k +1

now, 36k^2 +/-12k = 12(3k^2 +/-k)

3k^ +/-k is always even for odd and even k, thus
every p^2 > 3^2 can be written as 24n + 1 for some natural number n.
Thus f(n) is prime for infinite n.
Hence, (a) is false

We know that (24a + 1)^2 is always composite for infinite values of a.
also (24a + 1)^2 is of the form 24n + 1 where n is positive integer.
Hence, (b) is also false

Hence, choice (d) is the correct answer