# Help/Hint required.

Q1)Find the number of 4 digit numbers with distinct digits chosen from the set {0, 1, 2, 3, 4, 5} in which no two adjacent digits are even.

I'm not sure about the answer. I'm getting 60. There's a high chance I may be wrong.

Q2) Let a, b, c > 0. If $\dfrac 1 c , \dfrac 1 b \quad and \dfrac 1 a$ are in arithmetic progression, and if $a^2+b^2,b^2+c^2,c^2+a^2$ are in a geometric progression, prove that a=b=c

Somehow, I managed to get a=b=c. but I'm not really confident of my approach.

Q3) Let n be a positive integer such that 2n + 1 and 3n + 1 are both perfect squares. Show that 5n + 3 is a composite number.

I tried this

Let $2n+1=k^2$ and $3n+1=m^2$

$5n+3=m^2+k^2+1$

Then I used modulo to bash it out. am I on the right track? Because this doesn't seem to lead to the answer.

Thanks! Note by Mehul Arora
5 years, 10 months ago

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Q1) Let the odd digit be $O$ and even digit be $E$.So by product rule we can work out each case. So hereby I give the possible combinations:

$OEOO = 3\times 3 \times 2 \times 1 = 18 \\ OOEO = 3\times 2 \times 3 \times 1 =18 \\ EOOO = 2\times 3 \times 2 \times 1 = 12 \\ OOOE = 3\times 2 \times 1 \times 3 = 18 \\ EOOE = 2\times 3 \times 2 \times 2 = 24 \\ OEOE = 3\times 3 \times 2 \times 2 = 36 \\ EOEO = 2\times 3 \times 2 \times 2 = 24 \\ \Rightarrow 18+18+12+18+24+36+24 = \boxed{150}$

- 5 years, 10 months ago

Thanks Nihar! :D

- 5 years, 10 months ago

Saale dost ko thanks bolta hain :P

- 5 years, 10 months ago

Ahahahahahh xD

Chal nahi bolta thanks xD

Waapas liya thanks xD

- 5 years, 10 months ago

For the record, here are the numbers:

 1 3210,,5210,,1230,,5230,,1250,,3250,,3410,,5410,,1430,,5430,,1450,,3450,,3012,,5012,,3412,,5412,,1032,,5032,,1432,,5432,,1052,,1452,,3052,,3452,,3014,,5014,,3214,,5214,,1034,,5034,,1234,,5234,,1054,,3054,,1254,,3254,,5031,,5231,,5431,,3051,,3251,,3451,,5013,,5213,,5413,,1053,,1253,,1453,,3015,,3215,,3415,,1035,,1235,,1435,,2301,,2501,,2341,,2541,,2103,,2503,,2143,,2543,,2105,,2145,,2305,,2345,,4301,,4501,,4321,,4521,,4103,,4503,,4123,,4523,,4105,,4305,,4125,,4325,,5301,,3501,,5321,,3521,,5341,,3541,,5103,,1503,,5123,,1523,,5143,,1543,,3105,,3125,,3145,,1305,,1325,,1345,,2310,,2510,,2130,,2530,,2150,,2350,,2314,,2514,,2134,,2534,,2154,,2354,,4310,,4510,,4130,,4530,,4150,,4350,,4312,,4512,,4132,,4532,,4152,,4352,,2531,,2351,,2513,,2153,,2315,,2135,,4531,,4351,,4513,,4153,,4315,,4135,,5310,,3510,,5130,,1530,,3150,,1350,,5312,,3512,,5132,,1532,,3152,,1352,,5314,,3514,,5134,,1534,,3154,,1354, 

- 5 years, 10 months ago

Comp. Science Gawd is here. _/_

- 5 years, 10 months ago

For the 2nd one, We get that $b=2ac/a+c$ Substitute this in the second equation, $( {b}^2+{c}^2)/({a}^2+{b}^2)=({ c}^2+{a}^2)/({b}^2+{c}^2)$ By simplifying we get, a=b=c

- 5 years, 10 months ago

- 5 years, 10 months ago

Wait in Q1 when there asking for no two adjacent digits are even, doesn't that mean none of the digits are even ? O.O

- 5 years, 10 months ago

Not necessarily I guess. The placement of the digits could be OEOO

- 5 years, 10 months ago

Oh, so no two adjacent digits are both even?

- 5 years, 10 months ago

Exactly :P

- 5 years, 10 months ago

In that case I got 150 as an answer :(.

550 doesn't make sense because the maximum number of four digit numbers with distinct digits from the given set is $5 \cdot 5 \cdot 4 \cdot 3 = 300$ , which is less than your answer.

- 5 years, 10 months ago

Oh! I guess i mistook the question for a,b,c and d not necessarily being distinct. Lemme work this one out. Till then, please work the other questions :)

- 5 years, 10 months ago

Well I just took it in cases. You need at least one even and at most two evens.

So you just work from there.

- 5 years, 10 months ago

Oh okay, thanks! :D

- 5 years, 10 months ago

I got 60 as the answer. Could you please elaborate on how you got 150?

- 5 years, 10 months ago

Oh, I like question 3. The main way to show that a number is composite, is to factorize it and then show that the factors are not $\pm1$

Staff - 5 years, 10 months ago

1) 278

- 5 years, 10 months ago