# CSAT Sample Questions Working

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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**

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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**

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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**

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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

So the last digit of (n!)^4 is 0 for all n

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

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

So the last digit of (n!)^4 is 0 for all n

**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**.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

Just one thing: you said "Using eight 1 steps and 2 two steps, there will be 9 choose 2 ways" - did you mean "Using

**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. "?
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