How many three-digit positive integers exist, all of whose digits are 2's and/or 5's?
1 answer:
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Answer:
see explanation
Step-by-step explanation:
The common difference d of an arithmetic sequence is
d = -
=
-
Substitute in values and solve for k, that is
5k - 1 - 2k = 6k + 2 - (5k - 1)
3k - 1 = 6k + 2 - 5k + 1
3k - 1 = k + 3 ( subtract k from both sides )
2k - 1 = 3 ( add 1 to both sides )
2k = 4 ⇒ k = 2
--------------------------------------------------------
The n th term of an arithmetic sequence is
=
+ (n - 1)d
= 2k = 2 × 2 = 4 and
d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5
Hence
= 4 + (7 × 5) = 4 + 35 = 39
Answer:
There are 80 marks in total.
Step-by-step explanation:
Let the number of total marks be .
The percentage score of the student can be written as the ratio
.
However,
.
Equating the two:
.
Cross-multiply (that is: multiple both sides by , the product of the two denominators) to get
.
.
In other words, there are 80 marks in total.