Question-198684

Question Number 198684 by cortano12 last updated on 23/Oct/23

$$\:\cancel{\zeta} \\ $$

Answered by Red1ight last updated on 23/Oct/23

24,88 A=24×88=2112 P=24+88=112 2112⇒112=P

$$\mathrm{24},\mathrm{88} \\ $$$${A}=\mathrm{24}×\mathrm{88}=\mathrm{2112} \\ $$$${P}=\mathrm{24}+\mathrm{88}=\mathrm{112} \\ $$$$\mathrm{2112}\Rightarrow\mathrm{112}={P} \\ $$

Commented by mr W last updated on 23/Oct/23

$${how}\:{did}\:{you}\:{get}\:{these}\:{numbers}? \\ $$

Commented by Red1ight last updated on 23/Oct/23

$$\mathrm{by}\:\mathrm{trial}\:\mathrm{and}\:\mathrm{error} \\ $$

Commented by mr W last updated on 23/Oct/23

$${thanks}! \\ $$

Answered by Rasheed.Sindhi last updated on 23/Oct/23

Let a and b are 2-digit numbers such that: a+b+2000=ab ab−a−b=2000 a(b−1)−b+1=2000+1 (b−1)(a−1)=2001 (b−1)(a−1)=3×23×29 (b−1)(a−1)=23×87=69×29 b−1=23 ∧ a−1=87 b=24 ∧ a=88 ✓ b−1=29 ∧ a−1=69 b=30 ∧ a=70 ✓

$${Let}\:{a}\:{and}\:{b}\:{are}\:\mathrm{2}-{digit}\:{numbers} \\ $$$${such}\:{that}: \\ $$$${a}+{b}+\mathrm{2000}={ab} \\ $$$${ab}−{a}−{b}=\mathrm{2000} \\ $$$${a}\left({b}−\mathrm{1}\right)−{b}+\mathrm{1}=\mathrm{2000}+\mathrm{1} \\ $$$$\left({b}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)=\mathrm{2001} \\ $$$$\left({b}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)=\mathrm{3}×\mathrm{23}×\mathrm{29} \\ $$$$\left({b}−\mathrm{1}\right)\left({a}−\mathrm{1}\right)=\mathrm{23}×\mathrm{87}=\mathrm{69}×\mathrm{29} \\ $$$${b}−\mathrm{1}=\mathrm{23}\:\wedge\:{a}−\mathrm{1}=\mathrm{87} \\ $$$${b}=\mathrm{24}\:\wedge\:{a}=\mathrm{88}\:\:\:\checkmark \\ $$$${b}−\mathrm{1}=\mathrm{29}\:\wedge\:{a}−\mathrm{1}=\mathrm{69} \\ $$$${b}=\mathrm{30}\:\wedge\:{a}=\mathrm{70}\:\checkmark \\ $$

Commented by mr W last updated on 23/Oct/23

��

Commented by Rasheed.Sindhi last updated on 23/Oct/23

��

Commented by cortano12 last updated on 23/Oct/23

$$\mathrm{nice} \\ $$

Answered by mr W last updated on 23/Oct/23

say the two 2−digit numbers are x, y. xy=2pqr x+y=pqr xy−(x+y)=2000 x=((2000+y)/(y−1))≤99 ⇒y≥22, similarly x≥22 x=((2001)/(y−1))+1=((3×23×29)/(y−1))+1 y−1=23 ⇒y=24, x=88 y−1=29 ⇒y=30, x=70 (y−1=3×23=69 ⇒y=70, x=30) (y−1=3×29=87 ⇒y=88, x=24) so the two numbers are 24, 88 or 30, 70.

$${say}\:{the}\:{two}\:\mathrm{2}−{digit}\:{numbers}\:{are}\:{x},\:{y}. \\ $$$${xy}=\underline{\mathrm{2}{pqr}} \\ $$$${x}+{y}=\underline{{pqr}} \\ $$$${xy}−\left({x}+{y}\right)=\mathrm{2000} \\ $$$${x}=\frac{\mathrm{2000}+{y}}{{y}−\mathrm{1}}\leqslant\mathrm{99}\:\Rightarrow{y}\geqslant\mathrm{22},\:{similarly}\:{x}\geqslant\mathrm{22} \\ $$$${x}=\frac{\mathrm{2001}}{{y}−\mathrm{1}}+\mathrm{1}=\frac{\mathrm{3}×\mathrm{23}×\mathrm{29}}{{y}−\mathrm{1}}+\mathrm{1} \\ $$$${y}−\mathrm{1}=\mathrm{23}\:\Rightarrow{y}=\mathrm{24},\:{x}=\mathrm{88} \\ $$$${y}−\mathrm{1}=\mathrm{29}\:\Rightarrow{y}=\mathrm{30},\:{x}=\mathrm{70} \\ $$$$\left({y}−\mathrm{1}=\mathrm{3}×\mathrm{23}=\mathrm{69}\:\Rightarrow{y}=\mathrm{70},\:{x}=\mathrm{30}\right) \\ $$$$\left({y}−\mathrm{1}=\mathrm{3}×\mathrm{29}=\mathrm{87}\:\Rightarrow{y}=\mathrm{88},\:{x}=\mathrm{24}\right) \\ $$$${so}\:{the}\:{two}\:{numbers}\:{are}\: \\ $$$$\mathrm{24},\:\mathrm{88}\:{or}\:\mathrm{30},\:\mathrm{70}. \\ $$

Commented by cortano12 last updated on 23/Oct/23

$$\mathrm{great} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply