Question 1194944
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X = multiples of 5
X = {5, 10, 15, ...}


100 is the smallest three-digit number
100/7 = 14.2857 approximately which rounds up to 15
7*14 = 98 is too small
7*15 = 105 clears the threshold showing that 105 is the smallest three-digit number that's a multiple of 7


999 is the largest three-digit number
999/7 = 142.714 approximately 
142*7 = 994
143*7 = 1001 is too big
So 994 is the largest three-digit number that's a multiple of 7


Y = multiples of 7 that are three-digit numbers
Y = {105, 112, 119, ..., 987, 994}
Y = {7*15, 7*16, 7*17, ..., 7*141, 7*142}


Z = set of numbers in BOTH set X and set Y 
Z = multiples of 5 and multiples of 7 that are 3 digits
Z = multiples of 35 (aka 5*7) that are 3 digits
Z = {105, 140, 175, ..., 945, 980}
Z = {3*35, 4*35, 5*35, ..., 27*35, 28*35}


Notes: 
100/35 = 2.857 approximately to help locate the smallest three-digit multiple of 35 (round up)
999/35 = 28.54 approximately; this helps spot the largest three-digit multiple of 35 (round down)


Anyways, we have this set
Z = {3*35, 4*35, 5*35, ..., 27*35, 28*35}
Focus on the numbers up front
{3, 4, 5, ..., 27, 28}
We could count out the values and find there are 26 of them. 
That is a bit tedious. It's more efficient to use the rule below.


Rule: 
If a set of consecutive integers starts at 'a' and ends at 'b', then there are b-a+1 elements to that set
Example: {7,8,9,10} has b-a+1 = 10-7+1 = 4 elements
I'll leave the formal proof to the reader.


There are b-a+1 = 28-3+1 = 26 numbers in the set {3, 4, 5, ..., 27, 28}


Consequently, there are 26 numbers in the set 
{3*35, 4*35, 5*35, ..., 27*35, 28*35}
aka 
{105, 140, 175, ..., 945, 980}


Answer: <font color=red>26 numbers in common to both sets</font>


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Another approach:


As mentioned earlier, 105 is the smallest number in set Z.
This is the first term of the arithmetic sequence.
Let a = 105


d = 35 is the common difference because we're dealing with multiples of 35


The nth term of the arithmetic sequence is 
a(n) = a + d(n-1)
a(n) = 105 + 35(n-1)
a(n) = 105 + 35n-35
a(n) = 35n + 70


Plug in a(n) = 999 to reflect the largest three-digit number
Solve for n
a(n) = 35n + 70
999 = 35n + 70
35n = 999-70
35n = 929
n = 929/35
n = 26.542857 approximately


Plug in n = 26
a(n) = 35n + 70
a(26) = 35*26 + 70
a(26) = 980 which is the largest item of set Z


You should find that a(27) = 1015 which is too large.


It should be fairly self-explanatory at this point that there are 26 items in the set {a(1), a(2), a(3), ..., a(25), a(26)}
I.e. there are 26 items in the set {1, 2, 3, ..., 25, 26}
 

Answer: <font color=red>26 numbers in common to both sets</font>
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