1. If a box of 20 candy bars costs $39, what is the cost per candy bar?
2 answers:
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First, note that 
, which gives
so in fact we're dealing with the system
Now, 3, 23, and 31 are relatively prime, so we can use the Chinese remainder theorem. But before we do that, we need to rework and ultimately eliminate the coefficients of
.
From the second equivalence, it follows immediately is some multiple of 3; this is because 
is divisible by 3, but 3 doesn't divide 70, so it must divide .
For the third equivalence, we can write
. Then modulo 31, we have that 
, which leaves us with 
. We want a congruence of the form 
, and to get that we can multiply 
and 3 by the inverse of 8 modulo 31.
To find this inverse, we solve for
in the relation
by using the Euclidean algorithm, or making just making the observation that
, so 
. Distributing 4 across the third equivalence in our system gives 
.
So to recap, we now have
and we're ready to use the CRT.
As a starting point, let's take
It's clear that taken modulo 23, the latter two terms vanish and we have a remainder of 17, as desired. 63 is already a multiple of 3, but just to avoid doing more work later, let's multiply each term by 3 anyway to keep getting a remainder of 0:
Now multiply the first two terms by 31 to make sure they vanish when taken modulo 31. Meanwhile,
, but we want to get 12, so we would multiply this term by the inverse of 7 mod 31, and again by 12.
We can make a quick observation that
, so 
. Alternatively, you can use the Euclidean algorithm to find this inverse. Either way, we get
So we're told that
is our solution set, where
. We can "simplify" this slightly by finding the least positive residue modulo the product of our moduli, which would be
so we end up with
for integers
.
so in fact we're dealing with the system
Now, 3, 23, and 31 are relatively prime, so we can use the Chinese remainder theorem. But before we do that, we need to rework and ultimately eliminate the coefficients of
From the second equivalence, it follows immediately
For the third equivalence, we can write
To find this inverse, we solve for
by using the Euclidean algorithm, or making just making the observation that
So to recap, we now have
and we're ready to use the CRT.
As a starting point, let's take
It's clear that taken modulo 23, the latter two terms vanish and we have a remainder of 17, as desired. 63 is already a multiple of 3, but just to avoid doing more work later, let's multiply each term by 3 anyway to keep getting a remainder of 0:
Now multiply the first two terms by 31 to make sure they vanish when taken modulo 31. Meanwhile,
We can make a quick observation that
So we're told that
is our solution set, where
so we end up with
for integers
I attached a possible diagram for the above question as well as its corresponding choices.
Given choices were:
a) AC ≅ DBb) ∠ABD ≅ ∠CBDc) ΔAED≅ΔCEBd) ΔDCE≅ΔBCE
Based on the diagram, Choice C. <span>ΔAED≅ΔCEB is true.</span>
Given choices were:
a) AC ≅ DBb) ∠ABD ≅ ∠CBDc) ΔAED≅ΔCEBd) ΔDCE≅ΔBCE
Based on the diagram, Choice C. <span>ΔAED≅ΔCEB is true.</span>