Tourists And Buses

If there are 22 tourists in each bus, then one tourist would not have a seat.

If one bus isn't used, then there would be 23 tourists haven't a seat, and each bus can hold an equal number of tourists and accommodate all of them. So we could divide 23 on (b -1) (buses except one).

23 is a primary number, which can't be divide except on 1 or itself. (b -1) couldn't be one bus, because (22 tourists on it + 23 tourists haven't seats = more than the maximum capacity of it 32). So (b - 1) should be 23, there is no answer else.

b - 1 = 23

then, b = 24 buses

Now... to get t, we have to solve the equation: t = tourists in each bus \times b

There are two ways to solve this equation, which are: 1) "If there are 22 tourists in each bus, then one tourist would not have a seat." 2) "If one bus isn't used, then each bus can hold an equal number of tourists and accommodate all of them."

1) Means: t - 1 = 22 \times b = 22 \times 24 = 528 Then, t = 529

2) Means: t = 23 \times (b - 1) = 23 \times 23 = 529 Here... notice that: tourists in each bus would be 23, because they were 22, and when we deleted a bus, there would be 22 lost their bus, in addition to the tourist which hadn't a seat firstly. So this means: 23 tourists haven't a seat, and 23 buses have in each one 22 tourists, so each bus from the 23 buses should host one tourist which haven't seat,

finally it would reaches to 23 buses with 23 tourists in each bus.

So.. wow.. the finally step ッ

t + b = 529 + 24 = 553

553 is the correct solution

CONGRATS,
              you've been reached  (❀◕‿◕)

The number of tourists can be written as $22b + 1$

As there is an equal number of tourists in each bus if 1 less bus is used,

$22b +1 \equiv 0 \pmod{b-1}$

$22b -22 + 23 \equiv 0 \pmod{b-1}$

$23 \equiv 0 \pmod{b-1}$ , $b-1$ is a multiple of 23.

At 24 buses, there are 529 passengers. $t + b = \boxed{553}$

Assume that in the second case, the passenger number in each bus is k.So, we get 2 equations.

22b+1=t

t=kb-k.

Writing the value of t in the first equation, we get,

22b+1=kb-k

kb-22b=k+1

b(k-22)=k+1

$b=\frac{k+1}{k-22}$

Thus b is a whole number, the denominator is 1.

So k-22=1, and k=23.

So $b=\frac{23+1}{23-22}$ =24

So t=23 $\times$ 24-23=529.

So the answer is 529+24= $\boxed{553}$

Imagine that there are b buses with 22 tourists per bus. And there is one lone tourist that is outside. Removing the last bus, the tourists that are outside will be

$22 + 1 = 23$

According to the problem, these 23 passengers can be equally divided onto the remaining buses. Therefore we can say that the remaining buses are 23. Adding the last one would total up to 24 buses. Thus,

$t = 22b + 1$

$t = 22(24) + 1$

$t = 528 + 1$ $t = 529$

$t + b = 529 + 24 = \boxed{553}$

When we take away one bus we will need to accommodate its 22 tourists on all buses plus the one tourist= 23 tourists. As 23 is a prime number it will be divided only by itself! so we need 23 buses to accommodate the 23 tourists. Which means we should have had 24 buses in order to have 23 left after taking one bus away. 24x22= 528, 528+1= 529, 529/23= 23, t+b= 24+529= 553.

After not using one bus, no. of tourist left without a bus = 23 (22 from the bus + 1 without a bus) Those 23 must have been distributes one each on remaining buses, one each bus. 23 is a prime no. Thus the no. of tourist is 23x23 and no of buses =23 used + 1 empty t+b= 23x23 + 24

From the given condition, t = 22b + 1. Now, 22b + 1 should be divisible by b - 1 and the quotient must be less than 32. Make 22b + 1 as 22(b-1) + 23. This should be divisible by b -1 means 23 should be divisible by b -1. This means b - 1 must be 1 or 23. If it is 1, the quotient exceeds 32. So, b -1 is 23. This gives b = 24 and then t = 529. Thus we get t + b = 553. I enjoyed this problem!!

We have the two equations:

$22b=t-1$ $x(b-1)=t$

where $x$ is equal to the equal number of tourists on the bus.

Substituting $x(b-1)$ as $t$ into the first equation, we have

$22b=x(b-1)-1$

$\implies 22b=bx-x-1$

$\implies bx-22b=x+1$

$\implies b\times (x-22) = x+1$

where $x$ is an integer and $22 < x \le 32$ .

We also must have $x-22$ be a divisor of $x+1$ , since $b$ must be an integer.

The only value of $x$ that satisfies all conditions is $23$ , so we have $x=23$ . This gives $b=x+1=24$ and $t=23\times23=529$ , giving $t+b=24+529=\boxed{553}\ \blacksquare$ .

"If there are $22$ tourists in each bus, then one tourist would not have a seat", it means that: $t=22b+1$

"If one bus isn't used, then each bus can hold an equal number of tourists and accommodate all of them", it means that: $t=(b-1)\times k$ , where $k\leq 32$

So we have: $22b+1=k(b-1)\Longrightarrow 22b+1=kb-k\Longrightarrow k+1=b(k-22)$

Analysing $\pmod {k-22}$ we have: $k+1\equiv 0\pmod{k-22}\Longrightarrow 23\equiv0\pmod{k-22}$ (since $k\equiv22\pmod {k-22}$ ).

So, $k-22=23$ or $k-22=1$ , but in the first option, $k>32$ , absurd!!! So $k=23$

And then, $b=24$ and $t=529$ , so $b+t=553$

1. 22*b + 1 = t
2. (b-1)* X = t where X = equal number of tourists
3. 22 < X <=32
4. b = (1+X)/(X-22)
5. You need to find b (an integer) based on condition 3.

Suppose there are n buses.

Then there must be ((22*n)+1) tourists.

Also,if n-1 buses are used, then equal number of tourists will be present in each bus,i.e (no. of tourists)/(n-1) is a whole number.

Thus, (22n+1)/n-1 is a whole no.,W.

Let m=n-1.i.e n=m+1.

Thus the equation becomes

(22*(m+1)+1)/m=W

(22m+22+1)/m=W

22+23/m=W

Thus m=23 or 1 for W to be a whole no.

i.e n=24 or 2

If n=2,no.of tourists=45

i.e if one bus is left unused,then 45 tourists will travel in a single bus.This is not possible as maximum capacity of a bus is 32 tourists.

Thus n=24 and no.of tourists = 22*24 +1=529

Thus,t+b=24+529=553

22b + 1= t ............A
t/(b-1) = integer.
(22b + 1)/(b-1) = integer.
(22b -22 +23) (b-1) = integer
{(22b -22 )+23}/ (b-1) = integer
(22b -22 )/ (b-1) + 23/ (b-1) = integer
22 + 23/ (b-1) = integer.
23/ (b-1) = integer
only b= 24 can satisfy A.
t= 22 x 24 +1 = 529
b + t = 553

let say total buses is B

total tourists is T = 22 B + 1

then reduced one bus, so now total buses is B -1.

and number of tourists per bus = (22B + 1)/(B-1) = 22 + n where 1 <= n <= 10

then we'll have :

B = 23/n + 1

B will be integer if and only if n = 1

so, B = 24

T + B = 22 * 24 + 1 +24 = 553

"If there are tourists in each bus, then one tourist would not have a seat."

From this, we get 22b + 1 = t

"If one bus isn't used, then each bus can hold an equal number of tourists and accommodate all of them."

Let x = accommodated equal number of tourists when one bus isn't used

Therefore, x≦32 . Also, x(b-1) = t

Equating both equations, we get

22b + 1 = x(b-1)

b = (x+1)/(x-22)

Since b must be a whole number, (x+1)/(x-22) must also be a whole number.

To make (x+1)/(x-22) a whole number, the denominator may be = 1.

If x-22 = 1 , then x=23 (satisfying x≦32)

Hence, substituting x=23 , into the equations above, we get b=24 and t=529 .

Fianlly, b+t = 553

"each with a maximum capacity of 32 passengers" , and "If there are 22 tourists in each bus, then one tourist would not have a seat" , so:

22b+1=t (b,t∈N*) <=>b=(t-1)/22

and t<32b (#)

"If one bus isn't used, then each bus can hold an equal number of tourists and accommodate all of them" , so:

t=n(b-1) (n∈N*)

=> 22b+1=n(b-1)

=> b(n-22)=n+1

n=22 => impossible

n><22 => b=(n+1)/(n-22)=(n-22+23)/(n-22)=1+23/(n-22)

b∈N*=> n-22=23 or n-22=1 <=> n=45 or n=23 <=> b=2 (impossible (#) ) or b=24

=> (b;t)=(24;529)

=> *b+t=553 *

• 22b + 1 = t

• t mod b-1 = 0 => (22(b-1) + 23) mod b-1 = 0 => 23 mod b-1 = 0 => as 23 is prime number, only possible value of b = 24

• Dont know why 32 value is given for

from the first relation we get : t=22b+1 and from the second we get : b-1=t/b-1 by solving the two equations we get b=24 and t=529

22b+1=t; (b-1)23=t; b=24; t=529; t+b=553

t=22b+1.
If one bus is evacuated, there will be now 23 tourist left. Now we have to equally distribute 23 tourist in remaining b-1 buses. This is possible if the number of bus left is either 1 or 23 (1 can not be true as no. of tourist accommodated (45) will exceed as max capacity i.e 32 ). Hence total no of buses will be 24. and total no. of tourist will be 529.

t is one more than 22's multiple ---------> t = 22b+1

now if in 2nd case there are n tourists per bus,

t = n(b-1)

equating both we get

b(n-22) = n+1

Now we know three things

a) The number of tourists and buses are integers :P

b)The maximum value of n is 32

c) n has to be greater than 22 ( see above equation) - b has to be positive and an integer trying numbers from 23 to 32 we find 23 is the value of n

and hence b=23+1=24

and t = 22x24+1 = 529

∴ t+b=553

:) good day! hope you understood !

equations:

22n + 1 = Passengers x(n-1) = Passengers

22n + 1/n-1 = Integer 22n + 23 - 22/n-1 = Integer 22(n-1) + 23/n-1 = Integer n=24 (smallest)

therefore : no. of passengers + buses = 22n + 1 + n =553

For a long time, I felt wonderful while solving this. So, you have a number of tourists, call that "t", and you know that the exact number of buses, call it "b", contains 22 tourists however, 1 passenger doesn't has a seat so, here's the math: t=22b+1 now, you don't know the value of "b", but you know that the number of buses minus 1 gives you an exact of 23 because that one bus contained 22 passengers: b-1=23 the rest is easy: b=23+1=24 b=24 t=22 \times 24 +1 t=529 t+b=529+24=553. :)

at start if we consider that buses are running with their full capacity then the total no of passenger will be t=32 b but this is not the case,if there are 22 passengers in each bus then 1 would not get seat so no of passengers will be t-1=22 b now if 1 bus isn't used then all passengers can be accommodated equally in all remaining buses (b-1) as passengers are equally accommodated therefore t/(b-1)=(b-1) .i.e (22 b+1)/(b-1)=(b-1) solving this we'll get b=24 t=22 b+1 =529 if b-1 buses are used then t/(b-1)=529/23=23=(b-1) hence this is the solution t=529 b=24 t+b=553

22b+1=t. => 22(b-1)+23=t =>b-1=23 as (b-1)|22(b-1)+23 so b-1=23=>b=24, t=529. b+t=553