×

IS THIS POSSIBLE

P. Q .R =19(P+R)

HERE P, Q, R are distinct prime . find P, Q, R ????

NOTE - here P .Q .R means P multiplied by Q and R. means P. Q. R is not a 3 digit no.

Note by Jai Gupta
3 years, 5 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

ya you are correct but if they are not distinct primes then this is possible values are 2, 2 19

- 3 years, 5 months ago

If PQR = 19(P+R) then one of P, Q and R is 19. If Q=19 then P+R=PR which isn't possible for distinct primes. If P=19 (chosen without loss of generality), 19+R = QR. R must be odd, as 19+2 =21 which is not prime, so 19+R is even. As 19+R = QR, QR is even, Q is even, Q is 2. But the only solution to 19+R=2R is R=19 which, again, leads to P, Q and R not being distinct.

So... no. It isn't possible.

- 3 years, 5 months ago

×