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To practise my logical and mathematical skills, I attempted some of the sample CSAT questions, and was wondering if you could comment on my working, and if the answers are correct, and how I can improve. I will post my working for each question individually. Thanks 
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Q2: You decide to take the stairs and to climb either 1 or 2 stairs at a time, at any given time. In how many different ways can you get up a flight of 11 stairs? Prove your answer.
let f(n)be the number of ways for n stairs
for n > 2, the first step can either be 1 step or 2 steps, leaving a sequence of respectively n-1 or n-2 steps afterwards
therefore a general formula can be made that for n steps, the following steps will be a sequence of either n - 1 or n - 2 steps:
f(n) = f(n - 1) + f(n - 2)
Base cases:
f(0) = 0 [as clearly you cannot climb no steps]
f(1) = 1 [as clearly 1 step]
f(2) = 2 [as clearly 1 step then another, or 2 steps at once]
Therefore:
f(3) = f(2) + f(1) = 3
f(4) = f(3) + f(2) = 5
...
f(11) = 144
let f(n)be the number of ways for n stairs
for n > 2, the first step can either be 1 step or 2 steps, leaving a sequence of respectively n-1 or n-2 steps afterwards
therefore a general formula can be made that for n steps, the following steps will be a sequence of either n - 1 or n - 2 steps:
f(n) = f(n - 1) + f(n - 2)
Base cases:
f(0) = 0 [as clearly you cannot climb no steps]
f(1) = 1 [as clearly 1 step]
f(2) = 2 [as clearly 1 step then another, or 2 steps at once]
Therefore:
f(3) = f(2) + f(1) = 3
f(4) = f(3) + f(2) = 5
...
f(11) = 144
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Q3: What is the units digit of the number SIGMA [1337, n=1] (n!)^4?
First, work out the unit digits of each number 0-9 following (n!)^4
0: (0!)^4 = 0^4 = 0; unit digit is 0
1: (1!)^4 = 1^4 = 1; unit digit is 1
2: (2!)^4 = 2^4 = 16; unit digit is 6
3: (3!)^4 = 6^4 = 6(216); unit digit is 6 (6 * 6 = 36)
4: (4!)^4 = 24^4; unit digit is 6 (4^4 = 256)
5: (5!)^4 = 120^4 = 0; unit digit is 0 (0 ^ 4 = 0)
6: (6!)^4 = 720^4 = 0; unit digit is 0 (0 ^ 4 = 0)
... any factorial after 5 ends in a 0
9: unit digit is 0
The total of unit digits for numbers 1 to 10 = 1 + (3 * 6) = 19
The total of unit digits for numbers 1 to 7 = 1 + (3 * 6) = 19
Therefore the number 1337 can be treated as 1340, with 134 multiplied by the unit digit product
134 * 19 will result in a unit digit of 6 (4 * 9 = 36)
Therefore the answer is 6
First, work out the unit digits of each number 0-9 following (n!)^4
0: (0!)^4 = 0^4 = 0; unit digit is 0
1: (1!)^4 = 1^4 = 1; unit digit is 1
2: (2!)^4 = 2^4 = 16; unit digit is 6
3: (3!)^4 = 6^4 = 6(216); unit digit is 6 (6 * 6 = 36)
4: (4!)^4 = 24^4; unit digit is 6 (4^4 = 256)
5: (5!)^4 = 120^4 = 0; unit digit is 0 (0 ^ 4 = 0)
6: (6!)^4 = 720^4 = 0; unit digit is 0 (0 ^ 4 = 0)
... any factorial after 5 ends in a 0
9: unit digit is 0
The total of unit digits for numbers 1 to 10 = 1 + (3 * 6) = 19
The total of unit digits for numbers 1 to 7 = 1 + (3 * 6) = 19
Therefore the number 1337 can be treated as 1340, with 134 multiplied by the unit digit product
134 * 19 will result in a unit digit of 6 (4 * 9 = 36)
Therefore the answer is 6
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Q8: Let the function f be as defined below. What is the value of f(1337)?
function f(n):
if n < 1:
return 1
else if n is even:
return 1 + f(n-1)
else:
return 1 + f(n-3)
f(1337) = 1 + f(1334) = 1 + 1 + f(1333)
From this, we can tell that there is a loop of 2 + f(n-4)
1337 mod 4 = 1
Therefore we will be left with a number + f(1)
(1336 - 1) / 4 = 334
Therefore:
f(1337) = 2 * 334 + f(1) = 668 + f(1)
f(1) = 1 + f(-2) = 1 + 1 = 2
Therefore:
f(1337) = 668 + 2
f(1337) = 670
function f(n):
if n < 1:
return 1
else if n is even:
return 1 + f(n-1)
else:
return 1 + f(n-3)
f(1337) = 1 + f(1334) = 1 + 1 + f(1333)
From this, we can tell that there is a loop of 2 + f(n-4)
1337 mod 4 = 1
Therefore we will be left with a number + f(1)
(1336 - 1) / 4 = 334
Therefore:
f(1337) = 2 * 334 + f(1) = 668 + f(1)
f(1) = 1 + f(-2) = 1 + 1 = 2
Therefore:
f(1337) = 668 + 2
f(1337) = 670
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#5
For question 8 we did exactly the same thing.
For question 2, I got the same answer, but my working out was:
* Using all 1 steps, there is one way to climb the steps (11 choose 0)
* Using nine 1 steps and one 2 step, there are ten ways [depending on the position of the two step] (10 choose 1)
* Using seven 1 steps and two 2 steps, there will be 9 choose 2 ways.
So the total number of ways is:
(11 choose 0) + (10 choose 1) + ... + (6 choose 5)
[we can have a maximum of 5 two steps or we'll use more than 11 steps]
That sum evaluates to 1 + 10 + 36 + 56 + 35 + 6 = 144
But I think your method is more elegant.
For question 3, I disagree with your answer. I agree up to the point "any factorial after 5 ends in a 0". So we know that n! ends in a 0 for all n > 5. And any number ending in a 0 to the power of 4 will also end in a 0 (120^4 ends in a 0; so will 720^4, 5040^4 etc.). We don't need to consider groups of 10 because 11! does not end in a 1.
So the last digit of (n!)^4 is 0 for all n > 5. The sum from 1 to 1337 is the same as the sum from 1 to 4, which is simply (1 + 6 + 6 + 6) mod 10 = 19 mod 10 = 9.
For question 2, I got the same answer, but my working out was:
* Using all 1 steps, there is one way to climb the steps (11 choose 0)
* Using nine 1 steps and one 2 step, there are ten ways [depending on the position of the two step] (10 choose 1)
* Using seven 1 steps and two 2 steps, there will be 9 choose 2 ways.
So the total number of ways is:
(11 choose 0) + (10 choose 1) + ... + (6 choose 5)
[we can have a maximum of 5 two steps or we'll use more than 11 steps]
That sum evaluates to 1 + 10 + 36 + 56 + 35 + 6 = 144
But I think your method is more elegant.
For question 3, I disagree with your answer. I agree up to the point "any factorial after 5 ends in a 0". So we know that n! ends in a 0 for all n > 5. And any number ending in a 0 to the power of 4 will also end in a 0 (120^4 ends in a 0; so will 720^4, 5040^4 etc.). We don't need to consider groups of 10 because 11! does not end in a 1.
So the last digit of (n!)^4 is 0 for all n > 5. The sum from 1 to 1337 is the same as the sum from 1 to 4, which is simply (1 + 6 + 6 + 6) mod 10 = 19 mod 10 = 9.
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(Original post by ShatnersBassoon)
For question 8 we did exactly the same thing.
For question 2, I got the same answer, but my working out was:
* Using all 1 steps, there is one way to climb the steps (11 choose 0)
* Using nine 1 steps and 1 two step, there are ten ways [depending on the position of the two step] (10 choose 1)
* Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways.
So the total number of ways is:
(11 choose 0) + (10 choose 1) + ... + (6 choose 5)
[we can have a maximum of 5 two steps or we'll use more than 11 steps]
That sum evaluates to 1 + 10 + 36 + 56 + 35 + 6 = 144
But I think your method is more elegant.
For question 3, I disagree with your answer. I agree up to the point "any factorial after 5 ends in a 0". So we know that n! ends in a 0 for all n > 5. And any number ending in a 0 to the power of 4 will also end in a 0 (120^4 ends in a 0; so will 720^4, 5040^4 etc.). We don't need to consider groups of 10 because 11! does not end in a 1.
So the last digit of (n!)^4 is 0 for all n > 5. The sum from 1 to 1337 is the same as the sum from 1 to 4, which is simply (1 + 6 + 6 + 6) mod 10 = 19 mod 10 = 9.
For question 8 we did exactly the same thing.
For question 2, I got the same answer, but my working out was:
* Using all 1 steps, there is one way to climb the steps (11 choose 0)
* Using nine 1 steps and 1 two step, there are ten ways [depending on the position of the two step] (10 choose 1)
* Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways.
So the total number of ways is:
(11 choose 0) + (10 choose 1) + ... + (6 choose 5)
[we can have a maximum of 5 two steps or we'll use more than 11 steps]
That sum evaluates to 1 + 10 + 36 + 56 + 35 + 6 = 144
But I think your method is more elegant.
For question 3, I disagree with your answer. I agree up to the point "any factorial after 5 ends in a 0". So we know that n! ends in a 0 for all n > 5. And any number ending in a 0 to the power of 4 will also end in a 0 (120^4 ends in a 0; so will 720^4, 5040^4 etc.). We don't need to consider groups of 10 because 11! does not end in a 1.
So the last digit of (n!)^4 is 0 for all n > 5. The sum from 1 to 1337 is the same as the sum from 1 to 4, which is simply (1 + 6 + 6 + 6) mod 10 = 19 mod 10 = 9.
I like your method for q2, and there may be a way of making a sum for it, if we were to have it in terms of n steps. Just one thing: you said "Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways" - did you mean "Using seven 1 steps and 2 two steps, there will be 9 choose 2 ways. "?
Thanks
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#7
(Original post by some-student)
Just one thing: you said "Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways" - did you mean "Using seven 1 steps and 2 two steps, there will be 9 choose 2 ways. "?
Just one thing: you said "Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways" - did you mean "Using seven 1 steps and 2 two steps, there will be 9 choose 2 ways. "?
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