How many pairs of positive integers x, y exist such that HCF of x, y = 35 and sum of x and y = 1085?
How many pairs of positive integers x, y exist such that HCF of x, y = 35 and sum of x and y = 1085?
14
15
16
17
The Correct Option is B
Solution and Explanation
Let HCF of (x, y) be h. Then we can write x = h * a and y = h * b. Furthermore, note that HCF (a, b) = 1. This is a very important property. One that seems obvious when it is mentioned but a property a number of people overlook.
So, we can write x = 35a; y = 35b
x + y = 1085 => 35(a + b) = 1085. => (a + b) = 31. We need to find pairs of co-prime integers that add up to 31. (Another way of looking at it is to find out integers less than 31 those are co-prime with it or phi(31) as had mentioned. More on this wonderful function in another post).
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Concepts Used:
Acids and Bases
$P = - \frac{xx^T}{x^Tx}$Acid is any hydrogen-containing substance that is capable of donating a proton (hydrogen ion) to another substance. Base is an ion or molecule capable of accepting a hydrogen ion from acid.
Physical Properties of Acids and Bases
| Physical Properties | ACIDS | BASES |
| Taste | Sour | Bitter |
| Colour on Litmus paper | Turns blue litmus red | Turns red litmus blue |
| Ions produced on dissociation | H+ | OH- |
| pH | <7 (less than 7) | >7 (more than 7) |
| Strong acids | HCl, HNO3, H2SO4 | NaOH, KOH |
| Weak Acids | CH3COOH, H3PO4, H2CO3 | NH4OH |
Chemical Properties of Acids and Bases
| Type of Reaction | Acid | Bases |
| Reaction with Metals | Acid + Metal → Salt + Hydrogen gas (H2) E.g., Zn(s)+ dil. H2SO4 → ZnSO4 (Zinc Sulphate) + H2 | Base + Metal → Salt + Hydrogen gas (H2) E.g., 2NaOH +Zn → Na2ZnO2 (Sodium zincate) + H2 |
| Reaction with hydrogen carbonates (bicarbonate) and carbonates | Metal carbonate/Metal hydrogen carbonate + Acid → Salt + Carbon dioxide + Water E.g., HCl+NaOH → NaCl+ H2O 2. Na2CO3+ 2 HCl(aq) →2NaCl(aq)+ H2O(l) + CO2(g) 3. Na2CO3+ 2H2SO4(aq) →2Na2SO4(aq)+ H2O(l) + CO2(g) 4. NaHCO3+ HCl → NaCl+ H2O+ CO2 | Base+ Carbonate/ bicarbonate → No reaction |
| Neutralisation Reaction | Base + Acid → Salt + Water E.g., NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) | Base + Acid → Salt + Water E.g., CaO+ HCl (l) → CaCl2 (aq)+ H2O (l) |
| Reaction with Oxides | Metal oxide + Acid → Salt + Water E.g., CaO+ HCl (l) → CaCl2 (aq)+ H2O (l) | Non- Metallic oxide + Base → Salt + Water E.g., Ca(OH)2+ CO2 → CaCO3+ H2O |
| Dissolution in Water | Acid gives H+ ions in water. E.g., HCl → H+ + Cl- HCl + H2O → H3O+ + Cl– | Base gives OH- ions in water. |
Read more on Acids, Bases and Salts
Formulas Used:
Basic formula
$P = - \frac{xx^T}{x^Tx}$The a3 + b3 formula can be verified, by multiplying (a + b) (a2 - ab + b2) and see whether you get a3 + b3. The a3 + b3 formula or the difference of cubes formula is explained below:
a3 + b3 Formula = a3 + b3 = (a + b) (a2 - ab + b2)
You can remember these signs using the following trick.
Let us learn the a3 + b3 formula with a few solved examples.
Proof of a^3+ b^3 Formula
Let us see the verification of a cube plus b cube formula here. To prove or verify that a3 + b3 = (a + b) (a2 - ab + b2) we need to prove here LHS = RHS. Lets begin with the following steps.
LHS = a3 + b3
On Solving RHS side we get,
= (a + b) (a2 - ab + b2)
On multiplying the a and b separately with (a2 - ab + b2) we get
= a (a2 - ab + b2) + b(a2 - ab + b2)
= a3 - a2b + ab2 + a2b - ab2 + b3
= a3 - a2b + a2b + ab2- ab2 + b3
= a3 - 0 + 0 + b3
= a3 + b3
Hence proved, LHS = RHS
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