# HAST Exam Page (1 Viewer)

#### dem()sthenes

##### Member
No lol, that question isn't hard, it's just badly written.
I agree.

##### New Member
One of each shape works, but you could argue that the red heart is not an allowed shape in the first place. Your mirrored statement is just wrong. Take the middle column for instance.
Oh yeah. But it applies for every other column. And the shape can't "not be allowed" tbh.

#### hstb_

##### Member
Anyone else who just doesn't know wtf is going on and just worrying about trials and whatnot
lmaoo me

#### SaltSolutionPromo

##### New Member
Where can I find example HAST questions and/or past papers?

#### dem()sthenes

##### Member
Where can I find example HAST questions and/or past papers?
You can find it in their website. It's either pay or pirate.

#### Pahulbir Thind

##### New Member
POST MORE QUESTIONS PLEASE, ESPECIALLY MATHEMATICAL REASONING

#### Velocifire

##### Critical Hit
1) Find the degrees of each one

so u get 135 120 and 105 for each segment
Double check by adding to 360

then see if u can find the common factor between the three
seems 15 can fit perfectly

135/15 = 9
120/15 = 8
105/15 = 7

add them and u get 24 (a)
Fits the criteria of under 17 as well

a

#### D-BOSS

##### Active Member
This is AMC right?

#### Velocifire

##### Critical Hit
Angle sum to 360, that bit ain’t hard innit

just divide and use the 90 quarter as a reference point and add/subtract

@hifum also lucked out by using the 15 30 45 as Ages 4 8 and 12 which also got the correct answer but that’s just more like a guess and check rather than certainty with the angle proportion and taking a hcf or 15

#### Velocifire

##### Critical Hit
Also @hifum new problem the answer is 41 I think

#### Velocifire

##### Critical Hit
For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count
sets of 3 digits with all digits are odd, and they add to
a multiple of 3 (3, 6, 9, 12, 15, 18, 21, 24, or 27).
Those sets are:
{1,1,1} , {1,1,7} ,
{1,3,5} , {1,5,9} ,
{1,7,7} ,
{3,3,3} , {3,3,9} ,
{3,5,7} ,
{3,9,9} ,
{5,5,5} ,
{5,7,9} ,
{7,7,7} , and
{9,9,9} .

Among those
sets,
sets have 3 different digits,
Each of those
sets can be arranged in
different arrangements/permutations,
making
different three-digit numbers whose digits are all odd.

Among those
sets listed above,
there are also
sets made of just one single repeated digit,
and each one of those sets can be arranged just one way,
to form just one three-digit number,
so from them we can get another
three-digit numbers whose digits are all odd.

The remaining
of the
sets listed above contain only two different digits, one of them repeated.
Form each of those
sets, we can make
different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another
three-digit numbers whose digits are all odd.

That makes a total of
three-digit numbers divisible by 3, whose digits are all odd.

#### D-BOSS

##### Active Member
For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count
sets of 3 digits with all digits are odd, and they add to
a multiple of 3 (3, 6, 9, 12, 15, 18, 21, 24, or 27).
Those sets are:
{1,1,1} , {1,1,7} ,
{1,3,5} , {1,5,9} ,
{1,7,7} ,
{3,3,3} , {3,3,9} ,
{3,5,7} ,
{3,9,9} ,
{5,5,5} ,
{5,7,9} ,
{7,7,7} , and
{9,9,9} .

Among those
sets,
sets have 3 different digits,
Each of those
sets can be arranged in
different arrangements/permutations,
making
different three-digit numbers whose digits are all odd.

Among those
sets listed above,
there are also
sets made of just one single repeated digit,
and each one of those sets can be arranged just one way,
to form just one three-digit number,
so from them we can get another
three-digit numbers whose digits are all odd.

The remaining
of the
sets listed above contain only two different digits, one of them repeated.
Form each of those
sets, we can make
different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another
three-digit numbers whose digits are all odd.

That makes a total of
three-digit numbers divisible by 3, whose digits are all odd.
Wait I found that online lmao: https://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.978706.html

#### Velocifire

##### Critical Hit
Yeah that I did lol the other I didn’t
do you think I would have time to do that?

nice guys!!!

#### hifum

##### Active Member

this took me a while but i got the answer!

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