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ADHERETOLOGIC

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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.
 

Velocifire

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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
 

Velocifire

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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

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For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count
13
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
13
sets,
4
sets have 3 different digits,
Each of those
4
sets can be arranged in
3%2A2=6
different arrangements/permutations,
making
4%2A6=24
different three-digit numbers whose digits are all odd.

Among those
13
sets listed above,
there are also
5
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
5
three-digit numbers whose digits are all odd.

The remaining
4
of the
13
sets listed above contain only two different digits, one of them repeated.
Form each of those
4
sets, we can make
3
different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another
4%2A3=12
three-digit numbers whose digits are all odd.

That makes a total of
24%2B5%2B12=highlight%2841%29
three-digit numbers divisible by 3, whose digits are all odd.
 

D-BOSS

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For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count
13
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
13
sets,
4
sets have 3 different digits,
Each of those
4
sets can be arranged in
3%2A2=6
different arrangements/permutations,
making
4%2A6=24
different three-digit numbers whose digits are all odd.

Among those
13
sets listed above,
there are also
5
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
5
three-digit numbers whose digits are all odd.

The remaining
4
of the
13
sets listed above contain only two different digits, one of them repeated.
Form each of those
4
sets, we can make
3
different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another
4%2A3=12
three-digit numbers whose digits are all odd.

That makes a total of
24%2B5%2B12=highlight%2841%29
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
 

hifum

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Screen Shot 2020-07-28 at 8.32.34 pm.png

this took me a while but i got the answer!
 

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